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c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)

f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)

g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)

h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)

29 tháng 6 2021

Câu e đâu bạn?

6 tháng 7 2018

a)  \(1+\sqrt{3}+\sqrt{5}+\sqrt{15}\)

\(=\left(1+\sqrt{3}\right)+\sqrt{5}\left(1+\sqrt{3}\right)\)

\(=\left(1+\sqrt{3}\right)\left(1+\sqrt{5}\right)\)

b)  \(\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}\)

\(=\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{7}\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{7}\right)\)

c)  \(\sqrt{35}-\sqrt{15}+\sqrt{14}-\sqrt{6}\)

\(=\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{2}\right)\)

6 tháng 7 2018

e)  \(xy+y\sqrt{x}+\sqrt{x}+1\)

\(=y\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)

\(=\left(\sqrt{x}+1\right)\left(y\sqrt{x}+1\right)\)

g)  \(3+\sqrt{x}+9-x\)

\(=\left(3+\sqrt{x}\right)+\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)

\(=\left(3+\sqrt{x}\right)\left(4-\sqrt{x}\right)\)

11 tháng 4 2020

Đúng đề rồi á bạn

11 tháng 4 2020

Mà cảm ơn bạn nha

NV
1 tháng 8 2021

\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)

\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)

\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)

3 tháng 10 2021

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

bạn giải chi tiết giúp mk đc k ạ

 

AH
Akai Haruma
Giáo viên
16 tháng 8 2019

a)

\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)

\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)

b)

\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)

\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)

\(\Rightarrow B=0\)

c)

\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)

\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)

AH
Akai Haruma
Giáo viên
16 tháng 8 2019

d)

\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)

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

\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)

\(=\sqrt{2}.1^2=\sqrt{2}\)

e)

\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)

\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)

f)

\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)

2 tháng 8 2017

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)

18 tháng 9 2019

a)\(\frac{3\sqrt{6}-\sqrt{2}}{1-3\sqrt{3}}=\frac{3\sqrt{3}.\sqrt{2}-\sqrt{2}}{1-3\sqrt{3}}=\frac{\sqrt{2}.\left(3\sqrt{3}-1\right)}{-\left(3\sqrt{3}-1\right)}=-\sqrt{2}\)

b)\(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2.\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{5}}{2}\)

c)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}\)

d)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{5^2.6}-\sqrt{6^2.5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\sqrt{5}-\sqrt{30}.\sqrt{6}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}-\sqrt{6}}=\sqrt{30}\)

e)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{2^2.3}-\sqrt{3^2.2}}{\sqrt{6}}=\frac{\sqrt{6}.\sqrt{2}-\sqrt{6}.\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}}=\sqrt{2}-\sqrt{3}\)

f)\(\frac{6\sqrt{2}-4}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{16}}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{2}.2\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(6-2\sqrt{2}\right)}{\sqrt{2}}=6-2\sqrt{2}\)

g)\(\frac{6-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{36}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.2\sqrt{3}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.\left(2\sqrt{3}-5\right)}{\sqrt{3}}=2\sqrt{3}-5\)

19 tháng 9 2019

Cảm ơn bạn nha