Question

[nimbaoi12xyz]

Ai giúp em với ạ

Answer 1

[Yen Nhi]

a) \(A=\left(\frac{3}{x+3}+\frac{x}{3-x}-\frac{3x^2-9}{x^2-9}\right)\left(\frac{5}{x-2}+1\right)\) \(\left(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne2\end{cases}}\right)\)

\(=\frac{3\left(x-3\right)-x\left(x+3\right)-3x^2+9}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-2}\)

\(=\frac{3x-9-x^2-3x-3x^2+9}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-2}\)

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

b) \(\left(x+2\right)^2=4x^2\)

\(\Rightarrow\left(x+2\right)^2-\left(2x\right)^2=0\)

\(\Rightarrow\left(x+2-2x\right)\left(x+2+2x\right)=0\)

\(\Rightarrow\left(-x+2\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x+2=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\text{ (Loại)}\\x=\frac{-2}{3}\left(\text{(Thoả mãn)}\right)\end{cases}}\) 

Ta có: \(x=\frac{-2}{3}\)

\(\Rightarrow A=[-4\left(-\frac{2}{3}\right)^2]:\left(\frac{-2}{3}-3\right)\left(\frac{-2}{3}.2\right)=\frac{-4}{11}\)