giúp mới nà

giúp mình vs

ĐKXĐ: x\(x\ne\)1,-1

a) pt <=> \(\left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{x^2-1}=\frac{10}{9}\)

<=> \(\frac{4x^4}{\left(x^2-1\right)^2}-\frac{2x^2}{x^2-1}=\frac{10}{9}\)

Đặt: t=\(\frac{2x^2}{x^2-1}\)

Pt trở thành: \(t^2-t-\frac{10}{9}=0\)\(\Leftrightarrow9t^2-9t-10=0\)<=> \(\orbr{\begin{cases}t=-\frac{1}{3}\\t=\frac{5}{6}\end{cases}}\)

Nếu: \(\frac{2x^2}{x^2-1}=-\frac{1}{3}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{1}{7}}\\x=-\sqrt{\frac{1}{7}}\end{cases}\left(tm\right)}\)

Nếu: \(\frac{2x^2}{x^2-1}=\frac{5}{6}\)(vô nghiệm)

Vậy nghiệm là ...

http://vchat.vn/pictures/service/2017/02/iit1486637364.PNG

a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)

\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)

c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)