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12 tháng 7 2018
\(\text{a) }\sqrt{16-8\sqrt{3}}=\sqrt{12+4-4\sqrt{12}}=\sqrt{\left(\sqrt{12}-2\right)^2}=\sqrt{12}-2\)
\(\text{b) }\sqrt{38+12\sqrt{2}}=\sqrt{36+2+12\sqrt{2}}=\sqrt{\left(6+\sqrt{2}\right)^2}=6+\sqrt{2}\)
\(\text{c) }\sqrt{22+12\sqrt{2}}=\sqrt{18+2+4\sqrt{18}}=\sqrt{\left(\sqrt{18}+\sqrt{2}\right)^2}=3\sqrt{2}+\sqrt{2}=4\sqrt{2}\)
\(\text{d) }\sqrt{17-12\sqrt{2}}=\sqrt{9+8-6\sqrt{8}}=\sqrt{\left(3-\sqrt{8}\right)^2}=3-\sqrt{8}\)
\(\text{e) }\sqrt{20-10\sqrt{3}}=\sqrt{15+5-2\sqrt{75}}=\sqrt{\left(\sqrt{15}-\sqrt{5}\right)^2}=\sqrt{15}-\sqrt{5}\)
CW
1 tháng 8 2017
tớ ko chép lại đề, kí hiệu nhé
(1) \(=\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{\left|\sqrt{6}+\sqrt{5}\right|^2}=\left(\sqrt{6}-\sqrt{5}\right)^2-\left(\sqrt{6}+\sqrt{5}\right)=1-2\sqrt{30}-\sqrt{6}-\sqrt{5}\)
ai ra đề mà để đáp án dài thế này mất thẩm mĩ quá!!!
(2) \(=\sqrt{\left|\sqrt{5}+\sqrt{3}\right|^2}-\sqrt{\left|\sqrt{5}-\sqrt{3}\right|^2}=\left(\sqrt{5}+\sqrt{3}\right)-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
(3) \(=\sqrt{\left|\sqrt{7}+2\right|^2}-\sqrt{\left|3-\sqrt{5}\right|^2}=\sqrt{7}+2-3+\sqrt{5}=\sqrt{7}+\sqrt{5}-1\)
lại thêm 1 phép tính không đẹp....
(4) \(=\sqrt{\left|3\sqrt{2}-2\right|^2}-\sqrt{\left|3\sqrt{2}+1\right|^2}=3\sqrt{2}-2-3\sqrt{2}-1=-3\)
(5) \(=\sqrt{\left|2\sqrt{3}-1\right|^2}+\sqrt{\left|2\sqrt{3}-3\right|^2}=2\sqrt{3}-1+2\sqrt{3}-3=4\sqrt{3}-4\)
kiểm tra lại kết quả nhé ^^! Cảm ơn!
NN
2
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 !
DM
19 tháng 7 2017
\(\sqrt{18-2.3\sqrt{2}.2+4}\) +\(\sqrt{4+2.2.\sqrt{2}+2}\) =\(\sqrt{\left(3\sqrt{2}-2\right)^2}\) +\(\sqrt{\left(2+\sqrt{2}\right)^2}\)
=\(3\sqrt{2}-2+2+\sqrt{2}\) =\(4\sqrt{2}\)
25 tháng 6 2019
a) \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
b) \(\sqrt{12-6\sqrt{3}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{9}-\sqrt{3}\right)^2}\)
\(=3-\sqrt{3}\)
c) \(\sqrt{6-4\sqrt{2}+\sqrt{22-12\sqrt{2}}}\)
\(=\sqrt{6-4\sqrt{2}+\sqrt{\left(\sqrt{18}-2\right)^2}}\)
\(=\sqrt{6-4\sqrt{2}+\sqrt{18}-2}\)
\(=\sqrt{4-4\sqrt{2}+3\sqrt{2}}\)
\(=\sqrt{4-\sqrt{2}}\)
17 tháng 7 2017
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
NC
25 tháng 8 2020
1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}-10-5\sqrt{10}\)
\(=-10\)
2) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=14-2\sqrt{21}-7+2\sqrt{21}\)
\(=7\)
3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)
6 tháng 7 2017
a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\)
b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)
c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
MV
17 tháng 7 2017
a,
\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1+\sqrt{3}+1\\ =2\sqrt{3}\)
b,
\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\cdot\sqrt{4}\cdot\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\sqrt{20}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{20+4\sqrt{20}+4}+\sqrt{5-4\sqrt{5}+4}\\ =\sqrt{\left(\sqrt{20}+4\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{20}+4\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{20}+4+\sqrt{5}-2\\ =2+2\sqrt{5}+\sqrt{5}\\ =2+3\sqrt{5}\)
c,
\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{4}\cdot\sqrt{2}}+\sqrt{9+2\cdot\sqrt{4}\cdot\sqrt{2}}\\ =\sqrt{17-6\sqrt{8}}+\sqrt{9+2\sqrt{8}}\\ =\sqrt{9-6\sqrt{8}+8}+\sqrt{8+2\sqrt{8}+1}\\ =\sqrt{\left(3-\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{8}+1\right)^2}\\ =\left|3-\sqrt{8}\right|+\left|\sqrt{8}+1\right|\\ =3-\sqrt{8}+\sqrt{8}+1\\ =4\)
d,
\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{9}\cdot\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\sqrt{18}}\\ =\sqrt{4-4\sqrt{2}+2}+\sqrt{18-4\sqrt{18}+4}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{18}-2\right)^2}\\ =\left|2-\sqrt{2}\right|+\left|\sqrt{18}-2\right|\\ =2-\sqrt{2}+\sqrt{18}-2\\ =-\sqrt{2}+\sqrt{18}\\ =-\sqrt{2}+3\sqrt{2}\\ =2\sqrt{2}\)
\(\sqrt{22-12\sqrt{2}}\)
\(=\sqrt{(3\sqrt{2}-2)^2}\)
\(=|3\sqrt{2}-2|\)
`=3\sqrt{2}-2`
\(\sqrt{22-12\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}-2\right)^2}\)
\(=\left|3\sqrt{2}-2\right|\)
\(=3\sqrt{2}-2\)