# Bài 8: Rút gọn biểu thức chứa căn bậc hai

Đề bài: Rút gọn biểu thức:

1. $\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{^{ }\frac{a^4}{x^4}-1}$

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

3. $\left(\frac{3}{\sqrt{1+x}}\sqrt{1-x}\right)$ : $\left(\frac{3}{\sqrt{1-x^2}}+1\right)$

4. $\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)$ : $\left(\frac{a}{\sqrt{ab+b}}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)$

5. $\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}$ + $\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}$ .$\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)$

Các bạn giúp tớ nhé, hứa sẽ tick, tớ cảm ơn!!!!

3 tháng 7 2019 lúc 21:49

1.

Đặt $\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}$

$\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}$

$=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}$

$=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}$

$=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}$

2.

$=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]$

$=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})$

$=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)=(\sqrt{a}+1)^2(\sqrt{a}-1)^2$

$=(a-1)^2$

3.

$=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}$

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3 tháng 7 2019 lúc 21:59

4. Bạn xem lại đề xem đã đúng chưa?

5.

$=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}$

$=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}$

$=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}$

$=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}$

$=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}$

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