Question

Rút gọn

\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}+2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}+1\right)}\)

Answer 1

\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}+2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}+1\right)}\)

\(=\dfrac{\left[\left(\sqrt{x^2+4}\right)^2-2^2\right]\left(x+\sqrt{x}+1\right)\left(\sqrt{\left(\sqrt{x}-1\right)}\right)^2}{x\left(x\sqrt{x}+1\right)}\)

\(=\dfrac{\left[\left(x^2+4\right)-4\right]\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x\left(x\sqrt{x}+1\right)}\)

\(=\dfrac{\left(x^2+4-4\right)\left(\sqrt{x}^3-1\right)}{x\left(\sqrt{x}^3+1\right)}\)

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

\(=\dfrac{x\left(\sqrt{x}^3-1\right)}{\sqrt{x}^3+1}\)