# Bài 1: Căn bậc hai

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

$\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}$

$2\sqrt{5}-3\sqrt{45}+\sqrt{500}$

$\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}$

$\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}$

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

$\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}$

$\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}$

9 tháng 7 2018 lúc 21:25

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

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

c) $2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}$

d) $\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}$

e) $\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{4-3}+5\sqrt{3}=-2\sqrt{3}+5\sqrt{3}=3\sqrt{3}$

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

g) $\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1$

h)$\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}$

Bình luận (0)

Các câu hỏi tương tự
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết
Xem chi tiết