Rút gọn \(A=\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)
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Y
30 tháng 5 2023
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
5 tháng 10 2021
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
AH
Akai Haruma
Giáo viên
5 tháng 10 2021
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
25 tháng 6 2023
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 7 2021
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)
21 tháng 7 2023
`a, (2 sqrt 3 + sqrt 5)sqrt 3 - sqrt 60`
`= 2 sqrt 3 . sqrt 3 + sqrt 5 . sqrt 3 - sqrt(4 . 15)`
`= 2 . 3 + sqrt 15 - 2 sqrt 15`.
`= 6 - sqrt 15`.
`b, (5 sqrt 2 + 2 sqrt 5)sqrt 5 - sqrt250`
`= 5 sqrt 2 . sqrt 5 + 2 sqrt 5 . sqrt 5 - sqrt(25.10)`
`= 5 sqrt 10 + 10 - 5 sqrt 10`
`= 10`.
H9
HT.Phong (9A5)
CTVHS
21 tháng 7 2023
5) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
\(=2\sqrt{3}\cdot\sqrt{3}+\sqrt{5}\cdot\sqrt{3}-\sqrt{2^2\cdot15}\)
\(=2\cdot3+\sqrt{15}-2\sqrt{15}\)
\(=6+\left(1-2\right)\sqrt{15}\)
\(=6-\sqrt{15}\)
6) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)
\(=5\sqrt{10}+2\cdot5-5\sqrt{10}\)
\(=\left(5-5\right)\sqrt{10}+10\)
\(=0+10\)
\(=10\)
25 tháng 7 2023
\(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\dfrac{\sqrt{15-10\sqrt{2}}-\sqrt{10}}{\sqrt{5}}\)
\(=\left(\dfrac{-\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}-2\right)\left(\dfrac{\sqrt{5}\left(3+\sqrt{5}\right)}{3+\sqrt{5}}-2\right)\dfrac{\sqrt{5}.(\sqrt{3-2\sqrt{2}}-\sqrt{2})}{\sqrt{5}}\)
\(=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right).(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2})\)
\(=-3.-1=3\)
25 tháng 7 2023
`((5-2 sqrt 5)/(2-sqrt5) - 2)((5+3 sqrt 5)/(3+sqrt 5) - 2)`
`= (5-2sqrt 5 - 4 + 2 sqrt 5)/(2 -sqrt 5) . (5+3 sqrt 5 - 6 - 2 sqrt 5)/(3 + sqrt 5)`
`= 1/(2-sqrt5) . (-1 + sqrt 5)/(3 + sqrt 5)`
`= (sqrt 5 - 1)/((sqrt 5 + 3)(2 - sqrt 5))`
`b, (sqrt(15-10sqrt2) - sqrt 10)/(sqrt 5)`
`= (sqrt(10 - 2 . sqrt 5.sqrt 10 + 5) - sqrt 10)/(sqrt 5)`
`= (sqrt 10 - sqrt 5 - sqrt 10)/(sqrt5)`
`= -1`
7 tháng 1
a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)
b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)
d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)
e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)
\(A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\sqrt{4}\)
\(=6+2.2\)
\(=10\).Mà ta thấy A<0 nên \(A=-\sqrt{10}\)