(-3).8/8.6 rút gọn

chịu

\(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left(x+8-x+2\right)^2\)

\(=10^2\)

\(=100\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

\(\dfrac{8-2x}{x^2+x-20}=-\dfrac{2\left(4-x\right)}{\left(4-x\right)\left(x+5\right)}=\dfrac{-2}{x+5}\)

Để biểu thức trên nhận giá trị dương khi 

\(x+5< 0\)do -2 < 0 

\(\Leftrightarrow x< -5\)

a: Đặt \(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)

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

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

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

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

b: Thay x=1/2 vào A, ta được:

\(A=\frac{\left\lbrack\left(\frac12\right)^2+4\cdot\frac12-4\right\rbrack\left(\frac12+1\right)}{2\left\lbrack\left(\frac12\right)^2+4\right\rbrack}=\frac{\left(\frac14+2-4\right)\cdot\frac32}{2\left(\frac14+4\right)}=\frac{\frac32\cdot\frac{-7}{4}}{2\cdot\frac{17}{4}}=\frac{-21}{8}:\frac{17}{2}=-\frac{21}{8}\cdot\frac{2}{17}=\frac{-21}{68}\)