Câu hỏi của Cíuuuuuuuuuu

Question

Thực hiện phép chia:[(3-x)5-7(x-3)4-4(x-3)2] : (x2-6x+9)

Lấp La Lấp Lánh · Accepted Answer

$\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40$Câu trả lời: $\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40$

Nguyễn Lê Phước Thịnh · Answer

$\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}$$=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}$$=-\left(x-3\right)^3-7\left(x-3\right)^2-4$Câu trả lời: $\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}$$=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}$$=-\left(x-3\right)^3-7\left(x-3\right)^2-4$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP