# Bài 2: Căn thức bậc hai và hằng đẳng thức căn bậc hai của bình phương

Rút gọn biểu thức

a) $\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}$

b)$\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}$

c)$\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}$

d) $\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}$

e)$\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y},x,y>0$

f)$\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

g)$\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}v\text{ới}a>0,a\ne3$

7 tháng 8 2017 lúc 15:34

$\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}$

$=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}$

$=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}$

= - 1

$\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}$

$=\dfrac{\sqrt{6+2\sqrt{5}}}{2}$

$=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}$

$=\dfrac{\sqrt{5}+1}{2}$

$\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}$

$=\dfrac{2\sqrt{2}+2}{\sqrt{3+2\sqrt{2}}}$

$=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}$

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

= 2

$\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}$

$=4+9+16+49$

= 78

Bình luận (2)
7 tháng 8 2017 lúc 15:50

$\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}$

$=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}$

$=\sqrt{x}-\sqrt{y}$

$\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

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

$\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}$

$=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}$

$\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}$

$=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}$

$=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1$

= 3

Ta có:

$\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}$

$=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}$

$=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}$

$=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}$

$\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}$

$=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}$

$=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}$

Bình luận (2)

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