Đáp án:
$B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}$ với $x≥0;x\neq1$
Giải thích các bước giải:
$B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{5}{1-\sqrt{x}}+\dfrac{4}{x-1}$ ĐK: $x≥0;x\neq1$
$B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$B=\dfrac{(\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{5(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$B=\dfrac{x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$B=\dfrac{x+7\sqrt{x}+6}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$B=\dfrac{(\sqrt{x}+1)(\sqrt{x}+6)}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}$
Vậy $B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}$ với $x≥0;x\neq1$
