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NV
20 tháng 9 2019

\(\sqrt{7}-\sqrt{5}=\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{\sqrt{7}+\sqrt{5}}=\frac{2}{\sqrt{7}+\sqrt{5}}\)

\(\sqrt{5}-\sqrt{3}=\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}\)

Do \(\sqrt{7}>\sqrt{3}\Rightarrow\sqrt{7}+\sqrt{5}>\sqrt{5}+\sqrt{3}\Rightarrow\frac{2}{\sqrt{7}+\sqrt{5}}< \frac{2}{\sqrt{5}+\sqrt{3}}\)

\(\Rightarrow\sqrt{7}-\sqrt{5}< \sqrt{5}-\sqrt{3}\)

b/ \(\frac{1}{3}\sqrt{6}=\sqrt{\frac{2}{3}}\) ; \(6\sqrt{\frac{1}{3}}=2\sqrt{3}=\sqrt{12}\)

\(12>\frac{2}{3}\Rightarrow\sqrt{12}>\sqrt{\frac{2}{3}}\Rightarrow\frac{1}{3}\sqrt{6}< 6\sqrt{\frac{1}{3}}\)

23 tháng 4 2021

a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)

c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)

b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)

d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)

28 tháng 5 2021

a) Ta có: 33=32.3=9.3=27

Vì 27>12 nên 33>12

Vậy 33>12.
b) Ta có: 35=32.5=45

7=72=49

Vì 49>45 nên 7>35

Vậy 7>35.

c) Ta có: 1351=(13)2.51=519

15150=(15)2.150=15025=6=6.99=549

Vì 549>519 nên 1351<15150

Vậy 1351<15150.

d) Ta có: 126=(12)2.6=64

=32=3.12=3.12

Vì 3.12<612 nên 12.6<612

Vậy 126<612.

3 tháng 7 2019

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)

3 tháng 7 2019

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)

15 tháng 6 2017

Bài rút gọn 

\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)

\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)

Bài gpt:

\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)

Đk:\(-1\le x\le3\)

\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)

Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm

Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)

18 tháng 8 2016

a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

18 tháng 8 2016

a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)