Question

b=x(1-x)^2)/1+x^2 / [(1-x^2/1-x + x)(1+x^2/1+x - x)] a) rút ngọn b. b) cmb>0 với mọi x>0

Answer 1

Bạn gõ đề ở khung \(\Sigma\) cho đề rõ hơn nhé !

Answer 2

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

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

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

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

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

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

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

b: Đề này sai rồi bạn ,lỡ x=2 thì nó nhỏ hơn 0 á bạn