Question

Rút gọn:

B= \(\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right).\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)

Answer 1

điều kiện xác định : \(x>0;x\ne1\)

ta có : \(B=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)

\(\Leftrightarrow B=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)

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

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