Question

[Moon Chan lầy]

Mọi người ơi giải hộ mình các bài trên với ạ , mình cần gấp lắm ạ , cảm ơn mọi người!

Answer 1

[Nguyễn Minh Đăng]

Bài 1:

a) đk: \(x\ne\pm2\)

b) Ta có:

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

\(A=\left[\frac{1}{2-x}-\frac{3x}{\left(2-x\right)\left(2+x\right)}-\frac{2}{2+x}\right]\div\frac{x^2+4+4-x^2}{\left(2-x\right)\left(2+x\right)}\)

\(A=\frac{2+x-3x-2\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}\div\frac{8}{\left(2-x\right)\left(2+x\right)}\)

\(A=\frac{2-2x-4+2x}{\left(2-x\right)\left(2+x\right)}\cdot\frac{\left(2-x\right)\left(2+x\right)}{8}\)

\(A=\frac{-2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{\left(2-x\right)\left(2+x\right)}{8}=-\frac{1}{4}\)

=> đpcm

Answer 2

[Nguyễn Minh Đăng]

Bài 2: 

a) đk: \(x\ne\left\{-3;0;3\right\}\)

b) Ta có:

\(B=\left[\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right]\div\frac{3x^2}{x+3}\)

\(B=\left[\frac{-x^2-3x-9}{\left(x+3\right)^2}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)

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

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

\(B=-\frac{3}{x\left(x+3\right)}\)

c) Khi B = 1/2 thì: \(-\frac{3}{x\left(x+3\right)}=\frac{1}{2}\)

\(\Leftrightarrow x^2+3x=-6\Leftrightarrow x^2+3x+6=0\)

\(\Leftrightarrow\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{15}{4}=0\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2=-\frac{15}{4}\left(ktm\right)\)