Câu hỏi của Nguyễn Ngọc Châu Anh

Question

($\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{2011. 2012}$ ) x = 2011

Nguyễn Lê Phước Thịnh · Accepted Answer

$\Leftrightarrow x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)=2011$$\Leftrightarrow x\cdot\dfrac{2011}{2012}=2011$hay x=2012Câu trả lời: $\Leftrightarrow x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)=2011$$\Leftrightarrow x\cdot\dfrac{2011}{2012}=2011$hay x=2012

Nguyễn Ngọc Huy Toàn · Answer

$\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)x=2011$$\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)x=2011$$\left(\dfrac{1}{1}-\dfrac{1}{2012}\right)x=2011$$\dfrac{2011}{2012}x=2011$$x=2012$Câu trả lời: $\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)x=2011$$\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)x=2011$$\left(\dfrac{1}{1}-\dfrac{1}{2012}\right)x=2011$$\dfrac{2011}{2012}x=2011$$x=2012$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP