Question

\(P=\dfrac{10\sqrt{x}}{x+3\sqrt{x}-4}-\dfrac{2\sqrt{x}-3}{\sqrt{x}+4}+\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\) rút gọn

Answer 1

Lời giải:

ĐK: \(x\geq 0; x\neq 1\)

Ta có:

\(P=\frac{10\sqrt{x}}{x+4\sqrt{x}-\sqrt{x}-4}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}\)

\(=\frac{10\sqrt{x}}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{2x-5\sqrt{x}+3}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{x+5\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+4)}\)

\(=\frac{10\sqrt{x}-2x+5\sqrt{x}-3-x-5\sqrt{x}-4}{(\sqrt{x}+4)(\sqrt{x}-1)}\)

\(=\frac{10\sqrt{x}-3x-7}{(\sqrt{x}-1)(\sqrt{x}+4)}=\frac{(3\sqrt{x}-7)(1-\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+4)}\)

\(=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)