Question

Cho biểu thức:A=(\(\frac{x+1}{x-1}-\frac{x-1}{x+1}\))\(\div\)\(\frac{2x}{5x-5}\) với \(x\ne0\) và \(x\ne\pm1\).

a) Tính giá trị biểu thức.

B) Tìm x, để B=2

Answer 1

a)    A = ( \(\frac{x+1}{x-1}\)\(-\)\(\frac{x-1}{x+1}\))  \(\div\)\(\frac{2x}{5x-5}\)

= ( \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)\(-\)\(\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\))  \(\div\)\(\frac{2x}{5x-5}\)

= \(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)\(\div\)\(\frac{2x}{5x-5}\)

= \(\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(\times\)\(\frac{5\left(x-1\right)}{2x}\)

= \(\frac{4x}{\left(x-1\right)\left(x+1\right)}\)\(\times\)\(\frac{5\left(x-1\right)}{2x}\)

= \(\frac{10}{x+1}\)