# Biến đối đơn giản biểu thức chứa căn bậc hai

Thiếu tướng -
25 tháng 10 2020 lúc 19:51

a) Ta có: $\frac{6}{\sqrt{2}-\sqrt{3}+3}$

$=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}+3\right)\left(\sqrt{2}-\sqrt{3}-3\right)}$

$=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}$

$=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}$

$=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-2\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}$

$=\frac{3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}+\sqrt{3}\right)}{-\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}$

$=\frac{3\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}$

b) Ta có: $\left(\frac{4}{\sqrt{5}+1}-\frac{4}{\sqrt{5}-1}\right):\sqrt{3+2\sqrt{2}}$

$=\left(\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\right):\sqrt{2+2\cdot\sqrt{2}\cdot1+1}$

$=\left(\frac{4\left(\sqrt{5}-1\right)}{4}-\frac{4\left(\sqrt{5}+1\right)}{4}\right):\sqrt{\left(\sqrt{2}+1\right)^2}$

$=\left(\sqrt{5}-1-\sqrt{5}-1\right):\left|\sqrt{2}+1\right|$

$=-\frac{2}{\sqrt{2}+1}$(Vì $\sqrt{2}+1>0$)

$=-\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}$

$=-2\left(\sqrt{2}-1\right)$

$=-2\sqrt{2}+2$

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