\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
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Giải :
1) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
2) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
3) \(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
4) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)
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\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)
\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)
\(=2\sqrt{3}\)
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\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+5}\right):2\sqrt{5}\)
\(=\left(5\cdot\frac{\sqrt{1}}{\sqrt{5}}+\frac{1}{2}\sqrt{4.5}-\frac{5}{4}\sqrt{\frac{4+25}{5}}\right)\cdot\frac{1}{2\sqrt{5}}\)
\(=\left(\frac{5\sqrt{5}}{\sqrt{5}.\sqrt{5}}+\frac{1}{2}.2\sqrt{5}-\frac{5}{4}\sqrt{\frac{29}{5}}\right)\cdot\frac{\sqrt{5}}{2\cdot\sqrt{5}\cdot\sqrt{5}}\)
\(=\left(\frac{5\sqrt{5}}{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}}{\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(\sqrt{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}\sqrt{5}}{\sqrt{5}\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(2\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{145}}{5}\right)\cdot\frac{\sqrt{5}}{10}\)
\(=\left(2\sqrt{5}-\frac{\sqrt{145}}{4}\right)\cdot\frac{\sqrt{5}}{10}\)
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So sánh các số sau:
a = \(\frac{35}{49}\)b = \(\sqrt{\frac{5^2}{7^2}}\)c = \(\frac{\sqrt{5^2+\sqrt{35^2}}}{\sqrt{7^2}+\sqrt{49^2}}\)d = \(\frac{\sqrt{5^2-\sqrt{35^2}}}{\sqrt{7^2}-\sqrt{49^2}}\)
So sánh các số sau:
a = 3549 b = √5272 c = √52+√352√72+√492 d = √52−√352√72−√492
=> A < B
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bai nay minh chua hoc den nen khong the giai
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tính giá trị \(\frac{\sqrt{7}+7}{\sqrt{7}+1}-\frac{\sqrt{7}-\sqrt{14}}{\sqrt{2}-1}+\frac{2\sqrt{35}-2\sqrt{7}}{1-\sqrt{5}}\)
\(\frac{\sqrt{7}+7}{\sqrt{7}+1}-\frac{\sqrt{7}-\sqrt{14}}{\sqrt{2}-1}+\frac{2\sqrt{35}-2\sqrt{7}}{1-\sqrt{5}}\)
\(=\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{\sqrt{7}+1}-\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{\sqrt{2}-1}+\frac{2\sqrt{7}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\)
\(=\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{\sqrt{7}+1}+\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\frac{2\sqrt{7}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\)
\(=\sqrt{7}+\sqrt{7}-2\sqrt{7}\)
\(=0\)
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\(\frac{\sqrt{20}-\sqrt{12}+2}{\sqrt{35}-\sqrt{21}+\sqrt{7}}+\frac{5}{\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
So sánh
a)\(\sqrt{21}+\sqrt{5}\) và \(\sqrt{20}-\sqrt{6}\)
b)\(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}\) và \(\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
b) Ta có: \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{5+35}{7+49}=\frac{40}{56}=\frac{5}{7}\) (1)
Lại có: \(\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\frac{5-35}{7-49}=\frac{-30}{-42}=\frac{5}{7}\) (2)
Từ biểu thức (1) và biểu thức (2)
=> \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
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a) frac{2sqrt{5}-4sqrt{10}}{3sqrt{10}}b)frac{6sqrt{6}-2sqrt{12}+3-sqrt{2}}{2sqrt{6}+1}c)frac{sqrt{3left(3-sqrt{11}right)^2}}{sqrt{6}left(3-sqrt{11}right)}d)frac{5sqrt{7}-4sqrt{35}+7sqrt{5}}{sqrt{35}}e) left(sqrt{3}-1right)sqrt{2sqrt{19+8sqrt{3}-4}}g) sqrt{47+sqrt{5}}.sqrt{7-sqrt{2+sqrt{5}}.}sqrt{7+sqrt{2+sqrt{5}}}
Đọc tiếp
a) \(\frac{2\sqrt{5}-4\sqrt{10}}{3\sqrt{10}}\)
b)\(\frac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
c)\(\frac{\sqrt{3\left(3-\sqrt{11}\right)^2}}{\sqrt{6}\left(3-\sqrt{11}\right)}\)
d)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
e) \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}-4}}\)
g) \(\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+\sqrt{5}}.}\sqrt{7+\sqrt{2+\sqrt{5}}}\)
rut gon bieu thuc
C=\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)= ?
G=\(\sqrt{\frac{7}{5}}+\sqrt{\frac{5}{7}}-\frac{12}{35}\sqrt{35}\)= ?
P=\(\frac{a\sqrt{a}-1}{\sqrt{a}-1}\)=?
Rút Gọn Afrac{left(frac{1}{4}-frac{sqrt{2}}{7}+frac{3sqrt{2}}{35}right)frac{1}{25}}{left(frac{1}{10}+frac{3sqrt{2}}{35}-frac{sqrt{2}}{5}right)frac{5}{7}} Bfrac{3+sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{10}+sqrt{3+sqrt{5}}} Cfrac{3+sqrt{5}}{sqrt{2}+sqrt{3+sqrt{5}}}-frac{3-sqrt{5}}{sqrt{2}-sqrt{3-sqrt{5}}}
Đọc tiếp
Rút Gọn A=\(\frac{\left(\frac{1}{4}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\frac{1}{25}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{35}-\frac{\sqrt{2}}{5}\right)\frac{5}{7}}\)
B=\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
C=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)