\(=\frac{\left(5\sqrt{2}-2\sqrt{3}\right)^2+\left(5\sqrt{2}+2\sqrt{3}\right)^2}{\left(5\sqrt{2}-2\sqrt{3}\right)\left(5\sqrt{2}+2\sqrt{3}\right)}=\frac{50-20\sqrt{6}+12+50+20\sqrt{6}+12}{50-12}=\frac{124}{38}=\frac{62}{19}\)
rút gọn biểu thức:
1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2/ \(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3/ \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
4/ \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
giúp minh với cần gấp lắm
Q = \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
R = \(\frac{2}{7+4\sqrt{3}}+\frac{2}{7-4\sqrt{3}}\)
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
T = \(\frac{4}{1-\sqrt{3}}-\frac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
J = \(\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)\) . \(\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
Q = \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !
A) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
b) \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
c) \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
d) \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
e)\(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
g)\(\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
1 .
a)\(A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
b)\(B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)
c)C=\(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)
2 .
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b)\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c)C=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}}\)
d)D=(\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
* rút gọn biểu thức:
a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)
c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
Rút gọn
A= \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\) \(+\) \(\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{2}+\sqrt{3}}-\frac{5}{1+\sqrt{6}}\)
\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Đề bài: Khử mẫu của biểu thức dưới căn:
1. \(\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)
2. \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
3. \(\sqrt{\frac{3}{20}}\) + \(\sqrt{\frac{1}{60}}\) - 2\(\sqrt{\frac{1}{15}}\)
4. 2\(\sqrt{40\sqrt{12}}\) - 2\(\sqrt{\sqrt{75}}\) - 3\(\sqrt{5\sqrt{48}}\)
5. 2\(\sqrt{45\sqrt{3}}\) - 2\(\sqrt{20\sqrt{3}}\) - \(\sqrt{5\sqrt{48}}\)
Các bác giúp e vs, hứa sẽ tick, e cảm ơn!!!!!!
