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TN
1
19 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
19 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}\)
\(=\sqrt{2}\)
c: \(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
LN
1
25 tháng 6 2023
`a,\sqrt(3+2sqrt2)=\sqrt((sqrt2)^2+2.sqrt2 .1+1^2)=\sqrt((sqrt2+1)^2)=|sqrt2+1|=sqrt2+1`
`b,\sqrt(7+4sqrt3)=\sqrt((sqrt3)^2+2.\sqrt3 .2 +2^2)=\sqrt((sqrt3+2)^2)=|sqrt3+2|=sqrt3+2`
`c,sqrt(14-6sqrt5)=\sqrt((sqrt5)^2-2.\sqrt5 .3+3^2)=sqrt((sqrt5-3)^2)=|sqrt5-3|+3-sqrt5`
19 tháng 8 2021
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
Y
4 tháng 7 2023
\(a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\\ =\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\\ =\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\\ =2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\\ =2+2\sqrt{3}\)
\(b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\\ =\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\\ =\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\\ =-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\\ =-2+2\sqrt{2}-2\sqrt{2}+1\\ =-1\)
NT
1
13 tháng 12 2023
a: \(\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|3-\sqrt{5}\right|-\sqrt{5}\)
\(=3-\sqrt{5}-\sqrt{5}=3-2\sqrt{5}\)
b: \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=\left|4-2\sqrt{3}\right|\)
\(=4-2\sqrt{3}\)
13 tháng 5 2022
a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)
d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)
22 tháng 12 2023
a: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\dfrac{3}{2}\sqrt{2}+\dfrac{1}{2}\sqrt{2}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)
b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{18}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
d: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=-\sqrt[3]{27}=-3\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)