Question

1) cho biểu thức

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

a) tìm ĐKXĐ và rút gọn

b) tìm x để B=\(\dfrac{4}{3}\)

2) cho \(x^2+y^2=\dfrac{25xy}{12}\) và x>y>0

Tính giá trị biểu thức Q=\(\dfrac{x+y}{x-y}\)

Answer 1

Câu 1: 

a: ĐKXĐ: \(x\notin\left\{0;1;\dfrac{1}{2}\right\}\)

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

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

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

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

b: Để \(B=\dfrac{4}{3}\) thì \(\dfrac{x^2}{x^2+x+1}=\dfrac{4}{3}\)

\(\Leftrightarrow4x^2+4x+4-3x^2=0\)

=>x=-2(nhận)