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Chọn D.
x 4 + 8x = x( x 3 +8)= x( x 3 + 2 3 ) = x(x + 2)( x 2 − 2x + 4)
1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)
Điều kiện xác định: 9 - ( x + 5 ) 2 ≠ 0 .
Ta có:
Chọn đáp án A.
a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)
\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)
\(=\dfrac{x+1}{2x}\)
b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)
\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(8-x\right)}{x+2}\)
\(=\dfrac{x-8}{x+2}\)
c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)
\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)
\(=-2\)
d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
ĐKXĐ: \(x\neq0;x\neq-1\)
\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)
$---$
ĐKXĐ: \(x\neq y\)
\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)
\(=\dfrac{x-2}{x+2}\)
\(\dfrac{x^2-4x+4}{x^2-4}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)