`( x^2/(x^3 - 4x ) + 6/( 6 - 3x ) + 1/( 2 + x ) ) : ( x - 2 + ( 10 - x^2)/( x + 2 ) )` a, Rút gọn bt A b, tìm x để A = |A| c, Tìm x ∈Z để A ∈Z

Đáp án:

a)  $A=\dfrac{1}{2-x}$

b)  $x<2,x\ne 0,x\ne -2$

c)  $x\in \left\{ 1;3 \right\}$ 

Giải thích các bước giải:

$A=\left( \dfrac{{{x}^{2}}}{{{x}^{3}}-4x}+\dfrac{6}{6-3x}+\dfrac{1}{2+x} \right):\left( x-2+\dfrac{10-{{x}^{2}}}{x+2} \right)$

Điều kiện: $x\ne 0,x\ne \pm 2$

a)

$A=\left( \dfrac{{{x}^{2}}}{{{x}^{3}}-4x}+\dfrac{6}{6-3x}+\dfrac{1}{2+x} \right):\left( x-2+\dfrac{10-{{x}^{2}}}{x+2} \right)$

$A=\left[ \dfrac{{{x}^{2}}}{x\left( {{x}^{2}}-4 \right)}+\dfrac{6}{3\left( 2-x \right)}+\dfrac{1}{2+x} \right]:\left[ \dfrac{\left( x-2 \right)\left( x+2 \right)+10-{{x}^{2}}}{x+2} \right]$

$A=\left[ \dfrac{-x}{\left( 2-x \right)\left( 2+x \right)}+\dfrac{2}{2-x}+\dfrac{1}{2+x} \right]:\left[ \dfrac{{{x}^{2}}-4+10-{{x}^{2}}}{x+2} \right]$

$A=\dfrac{-x+2\left( 2+x \right)+\left( 2-x \right)}{\left( 2-x \right)\left( 2+x \right)}:\dfrac{6}{x+2}$

$A=\dfrac{6}{\left( 2-x \right)\left( 2+x \right)}\cdot \dfrac{x+2}{6}$

$A=\dfrac{1}{2-x}$

b)

Để $A=\left| A \right|$

Thì $A\ge 0$

$\Leftrightarrow \dfrac{1}{2-x}\ge 0$

$\Leftrightarrow 2-x>0$

$\Leftrightarrow x<2$

So với điều kiện, ta có $x<2,x\ne 0,x\ne -2$ thì $A=\left| A \right|$

c)

$A=\dfrac{1}{2-x}\in \mathbb{Z}$

$\Leftrightarrow 1\,\,\,\vdots \,\,\,\left( 2-x \right)$

$\Leftrightarrow 2-x\,\,\,\in \text{U}\left( 1 \right)=\left\{ 1;-1 \right\}$

$\Leftrightarrow -x\in \left\{ -1;-3 \right\}$

$\Leftrightarrow x\in \left\{ 1;3 \right\}$ (thỏa mãn điều kiện)