\(\left(\sqrt{7}-\sqrt{2}\right)^2+\sqrt{56}\)
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25 tháng 9 2021
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
25 tháng 9 2021
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
27 tháng 6 2021
`F=sqrt{(3-sqrt2)^2}+sqrt{(1-sqrt2)^2}``
`=3-sqrt2+sqrt2-1=2`
`G=sqrt{(5+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=5+sqrt7-(sqrt7-2)`
`=5+sqrt7-sqrt7+2=2`
`H=sqrt{(3-sqrt{10})^2}+sqrt{(2-sqrt{10})^2}`
`=sqrt{10}-3+sqrt{10}-2`
`=2\sqrt{10}-5`
27 tháng 6 2021
\(F=\left|3-\sqrt{2}\right|+\left|1-\sqrt{2}\right|=3-\sqrt{3}+\sqrt{2}-1=2\)
\(G=\left|5+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=5+\sqrt{7}-\sqrt{7}+2=7\)
\(H=\left|3-\sqrt{10}\right|+\left|2-\sqrt{10}\right|=\sqrt{10}-3+\sqrt{10}-2=2\sqrt{10}-5\)
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
\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)
\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)
\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)
\(=\sqrt{3}-1+2+\sqrt{3}=1+2\sqrt{3}\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
PT
1
NA
6 tháng 7 2015
\(A=\left(\sqrt{9-\sqrt{56}}-\sqrt{15+2\sqrt{56}}\right)\div5\sqrt{2}\)
\(A=\left[\left(\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\right)-\left(\sqrt{\left(\sqrt{8}+\sqrt{7}\right)^2}\right)\right]\div5\sqrt{2}\)
\(A=\left(\left|\sqrt{7}-\sqrt{2}\right|-\left|\sqrt{8}+\sqrt{7}\right|\right)\div5\sqrt{2}\)
\(A=\left(\sqrt{7}-\sqrt{2}-\sqrt{8}-\sqrt{7}\right)\div5\sqrt{2}\)
\(A=\left(-\sqrt{10}\right)\div5\sqrt{2}\)
\(A=\frac{-\sqrt{5}}{5}\)
\(\left(\sqrt{7}-\sqrt{2}\right)^2+\sqrt{56}=7+2-2\sqrt{14}+2\sqrt{14}=9\)
Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+\sqrt{56}\)
\(=9-2\sqrt{14}+2\sqrt{14}\)
=9