Đáp án:
Giải thích các bước giải:
ĐKXĐ : $x \geq 0 ; x\neq1$
$Q=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{5\sqrt{x}-1}{x-1}$
$Q=\dfrac{(2\sqrt{x}-1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{5\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$Q=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1+5\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$Q=\dfrac{2x+2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$Q=\dfrac{2\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$Q=\dfrac{2\sqrt{x}}{\sqrt{x}-1}$