\(\frac{\left(2x^3+2x\right)\left(x-2\right)^2}{\left(x^3-4x\ri...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

14 tháng 12 2016

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

\(=\frac{2x\left(x^2+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

\(=\frac{2\left(x^2+1\right)\left(x-2\right)}{\left(x+2\right)\left(x+1\right)}\)

Thay x=\(\frac{1}{2}\)

\(=\frac{2\left(\frac{1}{2}^2+1\right)\left(\frac{1}{2}-2\right)}{\left(\frac{1}{2}+2\right)\left(\frac{1}{2}+1\right)}\)

\(=-1\)

16 tháng 12 2017

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

\(=\)\(\frac{2\left(x-2\right)}{x+2}\)\(=\)\(\frac{2x-4}{x+2}\)

Tại   x = \(\frac{1}{2}\)thì:

             A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)

16 tháng 11 2018

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

\(A=\frac{\left(x^2+2x\right).\left(x^2-4x+4\right)}{\left(x^3-4x\right).\left(x+1\right)}=\frac{x^4-4x^3+4x^2+2x^3-8x^2+8x}{x^4+x^3-4x^2-4x}\)

\(A=\frac{x^4-2x^3-4x^2+8x}{x^4+x^3-4x^2-4x}=\frac{x^3.\left(x-2\right)-4x.\left(x-2\right)}{x^3.\left(x+1\right)-4x.\left(x+1\right)}=\frac{\left(x^3-4x\right).\left(x-2\right)}{\left(x^3-4x\right).\left(x+1\right)}=\frac{x-2}{x+1}\)

thay \(x=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}-2}{\frac{1}{2}+1}=\frac{-\frac{3}{2}}{\frac{3}{2}}=-1\)

Vậy A=-1 

29 tháng 11 2018

\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

\(=\frac{2\left(x-2\right)}{x+2}\)

Với \(x=\frac{1}{2}\)

\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)

b,Do x = -5; y = 10=> y = -2x

Thay y = -2x vào biểu thức ta được

\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)

\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)

\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)

Thay x = -5 là đc