Rút gọn biểu thức

Sai đề hay sao á

\(\left( {\frac{{{x^2}}}{{{x^3} - 4x}} + \frac{6}{{6 - 3x}} + \frac{1}{{x + 2}}} \right):\left( {x - 2 + \frac{{10 - {x^2}}}{{x + 2}}} \right)\)

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

\(=\left( {\frac{x}{{{x^2} - 4}} - \frac{6}{{3(x - 2)}} + \frac{1}{{x + 2}}} \right):\frac{6}{{x + 2}}\)

\(=\left( {\frac{x}{{(x + 2)(x - 2)}} - \frac{{2(x + 2)}}{{(x + 2)(x - 2)}} + \frac{{x - 2}}{{(x - 2)(x + 2)}}} \right):\frac{6}{{x + 2}}\)

\(= \frac{{x - 2x - 4 + x - 2}}{{(x + 2)(x - 2)}}.\frac{{x + 2}}{6}\)

\(= \frac{{ - 6}}{{(x + 2)(x - 2)}}.\frac{{x + 2}}{6} = \frac{1}{{2 - x}}\)

Đề như này mới ra đúng đáp án nhé

\(a) A=\frac{1}{2-x}\)

b) \(|x|=\frac{1}{2} \Leftrightarrow x=\frac{1}{2} và x=\frac{-1}{2}\)

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

\(x=-\frac{1}{2} \Rightarrow A=\frac{2}{5}\)

c) Để A<0 \(\Rightarrow \frac{1}{2-x}<0\)

\(\Leftrightarrow 2-x <0 \Leftrightarrow x>2\)