$3\div4=?$ $10000\times9-\left(2009-467\right)=?$ $98745-4524$

M

manh

15 tháng 3 2019 - olm

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}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$v

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Đ𝐚𝐧𝐧 𝐋ê

15 tháng 3 2019

ko đăng câu linh tinh

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ミ★ 🆂🆄🅽 ★彡

15 tháng 3 2019

I. Nội qui tham gia "Giúp tôi giải toán"

1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;

2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.

3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.

Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web.

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ND

Nguyễn Đăng Khôi

17 tháng 3 2020 - olm

$\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{4x}{2\left(x-3\right)\left(x+1\right)}=0$

$\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0$

$x^2-3x=0$

#Toán lớp 1

3

PH

Phan Hồng Lam

18 tháng 3 2020

đây không phải Toán Lớp lớp 1 đâu

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NT

nguyễn thùy chi

4 tháng 4 2021

đâu phải toán lớp 1

bạn chọn nhầm à

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NY

nguyen yen vi

24 tháng 1 2017 - olm

tìm x biết a)$x+2x+3x+4x+...+2015x=2016\times2017$b)$1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1008}-1}{4}$c)$\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x$d)tìm x nguyên biết $\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|=2500$e) tìm x nguyên...

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#Toán lớp 1

0

ZC

zZz Cool Kid_new zZz

9 tháng 8 2019 - olm

...

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#Toán lớp 1

6

ZC

zZz Cool Kid zZz

9 tháng 8 2019

$=\left(a+b-c\right)\left(a-b\right)^2$ nha !

P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.

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K

KAl(SO4)2·12H2O

9 tháng 8 2019

Giỏi quá à :3

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T

Tuấn

28 tháng 12 2015 - olm

ĐKXD: tự làm nhé ==' từ đề $\Rightarrow A=\frac{2\sqrt{y}-9}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}-\frac{\left(\sqrt{y}+3\right)\left(\sqrt{y}-3\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}+\frac{\left(2\sqrt{y}+1\right)\left(\sqrt{y}-2\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}$$\Rightarrow A=\frac{2\sqrt{y}-9-y+9+2y-3\sqrt{y}-2}{MC}$\(\Rightarrow...

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#Toán lớp 1

1

GS

Girl Shock

29 tháng 12 2015

lớp 1 chưa hok đâu bn tick nha

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LH

luyen hong dung

16 tháng 8 2018 - olm

Cho $x+y+z=\left(x-y\right)\left(y-z\right)\left(z-x\right)$

Chứng minh :$M=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3⋮81$

#Toán lớp 1

0

NK

Nguyễn Khang

8 tháng 9 2019 - olm

Khi thử đổi biến chứng minh Iran 96 và cái kết.... Mà chả biết lúc đổi biến có tính sai chỗ nào ko mà kết quả nó nhìn khủng khiếp quá:(Cho a, b, c là các số không âm thỏa mãn không có 2 số nào đồng thời bằng 0. Chứng minh rằng:$\left(ab+bc+ca\right)\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right)\ge\frac{9}{4}$Đặt $\left(a+b+c;ab+bc+ca;abc\right)=\left(3u;3v^2;w^3\right)$Cần...

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#Toán lớp 1

2

TB

Trần Bá Định

8 tháng 9 2019

WTF Toán Lớp 1

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NK

Nguyễn Khang

8 tháng 9 2019

thấy mẹ nhầm rồi, quy đồng quên nhân:(( mai rảnh check lại:((

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PT

pham thi thu trang

18 tháng 4 2018 - olm

...

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#Toán lớp 1

1

PG

Phan Gia Huy

9 tháng 1 2021

bạn trung học hay tiểu học vậy

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I

Incursion_03

26 tháng 5 2019 - olm

Đã bảo là liên hợp là ra mà đ tin hả Zũ ? -_- $x^3+\sqrt{\left(x+1\right)^3}=9x+8\left(x\ge-1\right)$$\Leftrightarrow\left(x^3+1\right)+\left(x+1\right)\sqrt{x+1}-9\left(x+1\right)=0$$\Leftrightarrow\left(x+1\right)\left(x^2-x+1+\sqrt{x+1}-9\right)=0$$\Leftrightarrow\orbr{\begin{cases}x=-1\left(Tm\right)\\x^2-x+\sqrt{x+1}-8=0\left(1\right)\end{cases}}$Giải $\left(1\right)\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)+\left(\sqrt{x+1}-2\right)=0$ ...

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#Toán lớp 1

3

T

T.Ps

26 tháng 5 2019

#)Trả lời :

Toán lớp 1 ak a ??? chắc 2 năm ns em còn k lm đc :v

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I

Incursion_03

26 tháng 5 2019

Bài 42 , Có $m=\sqrt[3]{4+\sqrt{80}}-\sqrt[3]{\sqrt{80}-4}$

$\Rightarrow m^3=4+\sqrt{80}-\sqrt{80}+4-3m\sqrt[3]{\left(4+\sqrt{80}\right)\left(\sqrt{80-4}\right)}$

$\Leftrightarrow m^3=8-3m\sqrt[3]{80-16}$

$\Leftrightarrow m^3=8-3m\sqrt[3]{64}$

$\Leftrightarrow m^3=8-12m$

$\Leftrightarrow m^3+12m-8=0$

Vì vậy m là nghiệm của pt $x^3+12x-8=0$

Bài 44, c, $D=\sqrt[3]{2+10\sqrt{\frac{1}{27}}}+\sqrt[3]{2-10\sqrt{\frac{1}{27}}}$

$\Rightarrow D^3=2+10\sqrt{\frac{1}{27}}+2-10\sqrt{\frac{1}{27}}+3D\sqrt[3]{\left(2+10\sqrt{\frac{1}{27}}\right)\left(2-10\sqrt{\frac{1}{27}}\right)}$

$\Leftrightarrow D^3=4+3D\sqrt[3]{4-\frac{100}{27}}$

$\Leftrightarrow D^3=4+3D\sqrt[3]{\frac{8}{27}}$

$\Leftrightarrow D^3=4+2D$

$\Leftrightarrow D^3-2D-4=0$

$\Leftrightarrow D^3-4D+2D-4=0$

$\Leftrightarrow D\left(D^2-4\right)+2\left(D-2\right)=0$

$\Leftrightarrow D\left(D-2\right)\left(D+2\right)+2\left(D-2\right)=0$

$\Leftrightarrow\left(D-2\right)\left[D\left(D+2\right)+2\right]=0$

$\Leftrightarrow\left(D-2\right)\left(D^2+2D+2\right)=0$

$\Leftrightarrow\left(D-2\right)\left[\left(D+1\right)^2+1\right]=0$

Vì [....] > 0 nên D - 2 = 0 <=> D = 2

Ý d làm tương tự nhá

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CD

Chicken Dinner

24 tháng 8 2021 - olm

1. $\frac{x^3-10x^2+25x}{x^2-5x}$$=0$ ( đkxđ: $x\ne0;5$)<=> $\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0$<=> $x-5=0$<=> vô no2. $A=$$\frac{2x^2-2}{x^3-x^2-4x+4}$$=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}$ ( a, đkxđ: $x\ne1;\pm2$)b, $A=0$<=> $\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0$<=> $x=-1$( TM) . Vậy $A=0\Leftrightarrow x=-1$3. ...

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#Toán lớp 1

3

DQ

Dương Quang Long

24 tháng 8 2021

lop 1kho the

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L

Lalisa

26 tháng 8 2021

Lớp 1 kiểu j vậy

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