# Ôn tập chương 1: Căn bậc hai. Căn bậc ba

19 tháng 9 lúc 12:05

a) $A=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x-3}{2x-x^2}$ (ĐK: $x\ne0;x\ne\pm2;x\ne3$)

$A=\left[-\dfrac{x+2}{x-2}+\dfrac{x-2}{x+2}-\dfrac{4x^2}{\left(x+2\right)\left(x-2\right)}\right]:\dfrac{x-3}{2x-x^2}$

$A=\left[\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{4x^2}{\left(x+2\right)\left(x-2\right)}\right]\cdot\dfrac{2x-x^2}{x-3}$

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

$A=\dfrac{-4x^2-8x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}$

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

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

$A=\dfrac{4x^2}{x-3}$

b) Ta có:

$\left|x\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.$

TH1: x=1

$A=\dfrac{4\cdot1^2}{1-3}=\dfrac{4}{-2}=-2$

TH2: x=-1

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

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19 tháng 9 lúc 12:19

a) $A=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x-3}{2x-x^2}\left(x\ne0;\pm2\right)$

$\Leftrightarrow A=\left[\dfrac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\right]-:\dfrac{x-3}{x\left(2-x\right)}$

$\Leftrightarrow A=\left[\dfrac{4+4x+x^2-\left(4-4x+x^2\right)+4x^2}{\left(2-x\right)\left(2+x\right)}\right]-:\dfrac{x-3}{x\left(2-x\right)}$

$\Leftrightarrow A=\left[\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right]:\dfrac{x-3}{x\left(2-x\right)}$

$\Leftrightarrow A=\left[\dfrac{4\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\right]:\dfrac{x-3}{x\left(2-x\right)}$

$\Leftrightarrow A=\dfrac{4}{2-x}.\dfrac{x\left(2-x\right)}{x-3}=\dfrac{4x}{x-3}$

b) $\left|x\right|=1$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.$

- Với $x=1:$

$A=\dfrac{4x}{x-3}=\dfrac{4.1}{1-3}=-2$

- Với $x=-1:$

$A=\dfrac{4x}{x-3}=\dfrac{4.\left(-1\right)}{-1-3}=1$

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