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 : \(a=\dfrac{35}{49};b=\sqrt{\dfrac{5^2}{7^2}};c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}};d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
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)
=> \(\frac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\frac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Rút gọn
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)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
d)\(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}\)
So sánh:
a, \(\frac{3\sqrt{7}+5\sqrt{2}}{\sqrt{5}}\) và \(\sqrt{35}+\sqrt{10}\)
b, \(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\) và \(\frac{1+\sqrt{5}}{2}\)
c, \(\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}\) và \(4\sqrt{2}\)
d, \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) và \(\sqrt{3}\)
\(\frac{3\sqrt{7}+5\sqrt{2}}{\sqrt{5}}=\frac{3}{5}\sqrt{35}+\sqrt{10}< \sqrt{35}+\sqrt{10}\)
\(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{6+2\sqrt{5}}}{2}=\frac{\sqrt{5}+1}{2}\)
\(\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\left(2+\sqrt{2}\right)^2+\left(2-\sqrt{2}\right)^2}{2}=\frac{12}{2}=6>4\sqrt{2}\) (do \(36>32\))
\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\frac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7}+1-\left(\sqrt{7}-1\right)}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}< \sqrt{3}\)
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=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
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) \(\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+\sqrt{5}}.}\sqrt{7+\sqrt{2+\sqrt{5}}}\)
rút gọn phân số sau:
A = \(\frac{1-\sqrt[1]{49}+\sqrt[1]{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\sqrt{\frac{64}{2}}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)