`B=\frac{3\sqrt{x} + 1}{x+2\sqrt{x}-3}` `-` `\frac{2}{\sqrt{x}+3}` Chứng minh `B=\frac{1}{\sqrt{x}-1}`
2 câu trả lời
$\text{B = $\dfrac{3\sqrt{x} + 1}{x + 2\sqrt{x} - 3}$ - $\dfrac{2}{\sqrt{x} + 3}$ Đkxđ:x ≥ 0; x $\neq$ 1}$
$\text{= $\dfrac{3\sqrt{x} + 1}{x - \sqrt{x} + 3\sqrt{x} - 3}$ - $\dfrac{2}{\sqrt{x} + 3}$}$
$\text{= $\dfrac{3\sqrt{x} + 1}{\sqrt{x}(\sqrt{x} - 1) + 3(\sqrt{x} - 1)}$ - $\dfrac{2}{\sqrt{x} + 3}$}$
$\text{= $\dfrac{3\sqrt{x} + 1}{(\sqrt{x} - 1)(\sqrt{x} + 3)}$ - $\dfrac{2}{\sqrt{x} + 3}$}$
$\text{= $\dfrac{(3\sqrt{x} + 1) - 2(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 3)}$}$
$\text{= $\dfrac{3\sqrt{x} + 1 - 2\sqrt{x} + 2}{(\sqrt{x} - 1)(\sqrt{x} + 3)}$}$
$\text{= $\dfrac{\sqrt{x} + 3}{(\sqrt{x} - 1)(\sqrt{x} + 3)}$}$
$\text{= $\dfrac{1}{\sqrt{x} - 1}$}$
$\text{Vậy B = $\dfrac{1}{\sqrt{x} - 1}$}$
$\textit{Ha1zzz}$
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