Question

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

a, Rút gọn biểu thức A 

b,Tính A biết x²-3x=0

c,Tĩm x để A\(\le\)-2 

 

 

 

các ace , cô bác rủ lòng thương vào giúp e vs  :((((

Answer 1

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

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

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

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

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

b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

+) x = 0 \(\Rightarrow A=\frac{-2.0+4}{0+2}=\frac{4}{2}=2\)

+) x = 3 \(\Rightarrow A=\frac{-2.3+4}{3+2}=\frac{-2}{5}\)

Answer 2

bạn giúp mk câu c vs