biểu thức \(\sqrt[3]{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\) có giá trị là bao nhiêu
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NB
6 tháng 3 2022
\(=\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{12+4\sqrt{6}+2+8\sqrt{3}+4\sqrt{2}+4-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{\left(2\sqrt{3}+\sqrt{2}\right)^2+4\left(2\sqrt{3}+\sqrt{2}\right)+4-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{\left(2\sqrt{3}+\sqrt{2}+2\right)^2-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(2\sqrt{3}+\sqrt{2}\right)=20\)
28 tháng 6 2021
`A=sqrt{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`
`=5-sqrt3+2-sqrt3`
`=7-2sqrt3`
`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`
`=3-sqrt2-(sqrt2-1)`
`=4-2sqrt2`
`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=3+sqrt7-(sqrt7-2)`
`=5`
`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`
`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`
`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`
`=sqrt3-1+2+sqrt3=1+2sqrt3`
28 tháng 6 2021
\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)
\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)
\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)
\(=\sqrt{3}-1+2+\sqrt{3}=1+2\sqrt{3}\)
RM
24 tháng 7 2017
a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}\)
\(=5+5-3-3\)
\(=4\)
b) \(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=\left(4+4\right).\frac{1}{8}-\frac{1}{9}\)
\(=8.\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
AT
2 tháng 7 2021
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
2 tháng 7 2021
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
1 tháng 7 2021
a, đặt \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(b,\)
\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
1 tháng 7 2021
a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1=2
b) Ta có: \(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=4\sqrt{7}\)
12 tháng 9 2021
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
12 tháng 9 2021
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6