1) Ta có: \(A=\dfrac{2x^2+2}{1-x^2}-\dfrac{1}{1+\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{2x^2+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(1+x\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(1+x\right)}{\left(1+x\right)\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\dfrac{\left(1+\sqrt{x}\right)\left(1+x\right)}{\left(1+x\right)\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(=\dfrac{2x^2+x\sqrt{x}+\sqrt{x}-x-1-1-x-\sqrt{x}-x\sqrt{x}}{\left(1+x\right)\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(=\dfrac{2x^2-2x-2}{\left(1+x\right)\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(=\dfrac{2x^2-2x-2}{1-x^2}\)
a) \(A=\dfrac{2x^2+2}{1-x^2}-\dfrac{1}{1+\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}\)
\(A=\dfrac{2x^2+2}{\left(1-x\right)\left(1+x\right)}-\dfrac{1-\sqrt{x}}{1-x}-\dfrac{1+\sqrt{x}}{1-x}\)
\(A=\dfrac{2x^2+2}{\left(1-x\right)\left(1+x\right)}+\dfrac{-2}{1-x}\)
\(A=\dfrac{2x^2+2}{\left(1-x\right)\left(1+x\right)}-\dfrac{2+2x}{\left(1-x\right)\left(1+x\right)}\)
\(A=\dfrac{2x\left(x-1\right)}{\left(1-x\right)\left(1+x\right)}=\dfrac{-2x}{1+x}\)
b) vì x ≥0 => -2.x≤0 và 1+x≥1
=> \(A\text{ }\text{≤}\dfrac{-2.0}{1+0}=0\) => A max =0 khi và chỉ khi x=0
