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25 tháng 5 2018

4 nha bn k mk nha

25 tháng 5 2018

Ta có:  \(\sqrt{22-12\sqrt{2}}-\)\(\sqrt{22+12\sqrt{2}}\)

            =\(\sqrt{18+2.2.3\sqrt{2}+4}\)\(-\sqrt{18-2.2.3\sqrt{2}+4}\)

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

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

            =  4

 Chúc bạn học tốt  !

1 tháng 10 2021

\(\sqrt{6-4\sqrt{2}}\)\(+\sqrt{22-12\sqrt{2}}\)

\(=\sqrt{4-4\sqrt{2}+2}\)\(+\sqrt{18-12\sqrt{2}+4}\)

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

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

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

\(=0+2\sqrt{2}\)\(=2\sqrt{2}\)

1 tháng 10 2021

\(\sqrt{17-12\sqrt{2}}\)\(+\sqrt{9+4\sqrt{2}}\)

\(=\sqrt{\left(3-2\sqrt{2}\right)^2}\)\(+\sqrt{\left(2\sqrt{2}+1\right)^2}\)

\(=\left|3-2\sqrt{2}\right|\)\(+\left|2\sqrt{2}+1\right|\)

\(=3-2\sqrt{2}\)\(+2\sqrt{2}+1\)

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

\(=4+0=4\)

23 tháng 6 2023

\(A=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)

\(=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+\sqrt{1^2}}-\sqrt{\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}+2^2}\)

\(=-2\sqrt{3}+3\sqrt{2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)

\(=-2\sqrt{3}+3\sqrt{2}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\)

\(=-2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-1-3\sqrt{2}-2\)

\(=-3\)

\(A=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2=-3\)

12 tháng 7 2018

\(\text{a) }\sqrt{16-8\sqrt{3}}=\sqrt{12+4-4\sqrt{12}}=\sqrt{\left(\sqrt{12}-2\right)^2}=\sqrt{12}-2\)

\(\text{b) }\sqrt{38+12\sqrt{2}}=\sqrt{36+2+12\sqrt{2}}=\sqrt{\left(6+\sqrt{2}\right)^2}=6+\sqrt{2}\)

\(\text{c) }\sqrt{22+12\sqrt{2}}=\sqrt{18+2+4\sqrt{18}}=\sqrt{\left(\sqrt{18}+\sqrt{2}\right)^2}=3\sqrt{2}+\sqrt{2}=4\sqrt{2}\)

\(\text{d) }\sqrt{17-12\sqrt{2}}=\sqrt{9+8-6\sqrt{8}}=\sqrt{\left(3-\sqrt{8}\right)^2}=3-\sqrt{8}\)

\(\text{e) }\sqrt{20-10\sqrt{3}}=\sqrt{15+5-2\sqrt{75}}=\sqrt{\left(\sqrt{15}-\sqrt{5}\right)^2}=\sqrt{15}-\sqrt{5}\)

19 tháng 3 2017

bằng 0,5 nhé 

16 tháng 6 2017

a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)

= \(6-\sqrt{15}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)

c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)

= \(7\)

d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)

23 tháng 4 2017

a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15

b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10

c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7

d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22

20 tháng 6 2021

\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)

\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{2}+1}\)

\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2}+1-\sqrt{2}+1=2\)