tính :
\(\sqrt{11-2\sqrt{30}}-\sqrt{11+2\sqrt{30}}\)
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N
neverexist_
27 tháng 12 2021
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Những câu hỏi liên quan
29 tháng 1 2018
Đặt \(A=\sqrt{11-2\sqrt{30}}-\sqrt{11+2\sqrt{30}}\)
\(\Leftrightarrow A^2=11-2\sqrt{30}+11+2\sqrt{30}-2\sqrt{\left(11-2\sqrt{30}\right)\left(11+2\sqrt{30}\right)}\)
\(\Leftrightarrow A^2=22-2\sqrt{11^2-\left(2\sqrt{30}\right)^2}\)
\(\Leftrightarrow A^2=22-2=20\)
\(\Leftrightarrow A=\pm\sqrt{20}\)
29 tháng 1 2018
Vì \(\sqrt{11-2\sqrt{30}}< \sqrt{11+2\sqrt{30}}\)
Nên A chỉ nhận giá trị \(-\sqrt{20}\)
2 tháng 10 2021
\(a,=\sqrt{\dfrac{81}{25}}=\dfrac{9}{5}\\ b,\approx6,39\\ c,=\sqrt{8,1\cdot20\cdot8}=\sqrt{81\cdot16}=\sqrt{81}\cdot\sqrt{16}=9\cdot4=36\\ d,=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\\ =\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}=2\sqrt{5}\)
2 tháng 10 2021
a) \(\sqrt{3\dfrac{6}{25}}=\sqrt{\dfrac{81}{25}}=\dfrac{9}{5}\)
b) \(\sqrt[3]{216}=6\)
c) \(\sqrt{8,1}.\sqrt{20}.\sqrt{8}=\dfrac{9\sqrt{10}}{10}.2\sqrt{5}.2\sqrt{2}=36\)
d) \(\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}=2\sqrt{5}\)
23 tháng 7 2021
1.
Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)
\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)
\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)
\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)
24 tháng 7 2021
\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)
\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)
\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)
\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)
ND
1
29 tháng 10 2017
\(\sqrt{11-2\sqrt{30}}-\sqrt{11+2\sqrt{30}}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\)
\(=\sqrt{6}-\sqrt{5}-\sqrt{6}-\sqrt{5}\)
\(=-2\sqrt{5}\)
NT
1
DD
Đoàn Đức Hà
Giáo viên
26 tháng 5 2021
\(A=\left(\sqrt{22}+7\sqrt{2}\right)\sqrt{30-7\sqrt{11}}\)
\(2A=\left(\sqrt{44}+7\sqrt{4}\right)\sqrt{60-2.7\sqrt{11}}\)
\(2A=\left(2\sqrt{11}+14\right)\sqrt{7^2-2.7\sqrt{11}+\left(\sqrt{11}\right)^2}\)
\(2A=\left(2\sqrt{11}+14\right)\sqrt{\left(7-\sqrt{11}\right)^2}\)
\(2A=\left(2\sqrt{11}+14\right)\left|7-\sqrt{11}\right|\)
\(2A=\left(2\sqrt{11}+14\right)\left(7-\sqrt{11}\right)\)
\(A=\left(7+\sqrt{11}\right)\left(7-\sqrt{11}\right)\)
\(A=49-11=38\)
TL
1
25 tháng 12 2018
\(D=\sqrt{6-2.\sqrt{5}.\sqrt{6}+5}-\sqrt{6+2.\sqrt{5}.\sqrt{6}+5}\)
\(D=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\)
\(D=|\sqrt{6}-\sqrt{5}|-|\sqrt{6}+\sqrt{5}|\)
\(D=\sqrt{6}-\sqrt{5}-\sqrt{6}-\sqrt{5}\)
\(D=-2\sqrt{5}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)
\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)
\(=2\sqrt{6}-1\)
18 tháng 7 2023
\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)