\(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{6}+2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
Những câu hỏi liên quan
rút gọn
a) \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
b) \(\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
c) \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
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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}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
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x^3left(sqrt[3]{5+2sqrt{6}}+sqrt[3]{5-2sqrt{6}}right)^3sqrt[3]{5+2sqrt{6}}^3+3sqrt[3]{left(5+2sqrt{6}right)^2}.sqrt[3]{5-2sqrt{6}}+3sqrt[3]{5+2sqrt{6}}.sqrt[3]{left(5-2sqrt{6}right)^2}+sqrt[3]{5-2sqrt{6}}^35+2sqrt{6}+3sqrt[3]{left(5+2sqrt{6}right)left(5-2sqrt{6}right)}.sqrt[3]{5+2sqrt{6}}+3sqrt[3]{left(5+2sqrt{6}right)left(5-2sqrt{6}right)}.sqrt[3]{5-2sqrt{6}}+5-2sqrt{6}5+5+3sqrt[3]{left(25-4.6right)}.sqrt[3]{5+2sqrt{6}}+3sqrt[3]{left(25-4.6right)}.sqrt[3]{5-2sqrt{6}}10+
3sqrt[3]{5+2sqrt{6}}+3sq...
Đọc tiếp
\(x^3=\left(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\right)^3=\sqrt[3]{5+2\sqrt{6}}^3\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)^2}.\sqrt[3]{5-2\sqrt{6}}+3\sqrt[3]{5+2\sqrt{6}}.\sqrt[3]{\left(5-2\sqrt{6}\right)^2}+\sqrt[3]{5-2\sqrt{6}}^3\)
\(=5+2\sqrt{6}+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5+2\sqrt{6}}\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5-2\sqrt{6}}+5-2\sqrt{6}\)
\(=5+5+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5-2\sqrt{6}}\)
\(=10+ 3\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{5-2\sqrt{6}}\)
p/s : có bạn hỏi nên mình đăng , các bạn đừng report nhé
Rút gọn :
a) left(sqrt{6}+sqrt{2}right).left(sqrt{3}-2right)left(sqrt{2+sqrt{3}}right)
b) sqrt{2}.left(sqrt{2-sqrt{3}}right).left(sqrt{3}+1right)
c) left(sqrt{10}-sqrt{6}right).left(sqrt{4-sqrt{15}}right)
d)left(sqrt{3}-sqrt{12}right).left(sqrt{5+2sqrt{6}}right)
e) sqrt{2-sqrt{3}}.left(sqrt{6}-sqrt{2}right).left(2+sqrt{3}right)
f) sqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
Đọc tiếp
Rút gọn :
a) \(\left(\sqrt{6}+\sqrt{2}\right).\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)\)
b) \(\sqrt{2}.\left(\sqrt{2-\sqrt{3}}\right).\left(\sqrt{3}+1\right)\)
c) \(\left(\sqrt{10}-\sqrt{6}\right).\left(\sqrt{4-\sqrt{15}}\right)\)
d)\(\left(\sqrt{3}-\sqrt{12}\right).\left(\sqrt{5+2\sqrt{6}}\right)\)
e) \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
f) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
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rút gọn A)\(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}\)
B) \(\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}\)
C) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
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Tính (rút gọn)
a) left(sqrt{3+sqrt{5}}right)left(sqrt{10}+sqrt{2}right)left(3-sqrt{5}right)
b) left(4+sqrt{15}right)left(sqrt{10-sqrt{6}}right)sqrt{4-}sqrt{15}
c) left(sqrt{6}+sqrt{2}right)left(sqrt{3}-2right)sqrt{sqrt{3+2}}
d) left(2sqrt{4+sqrt{6-2sqrt{5}}}right)left(sqrt{10}-sqrt{2}right)
f)sqrt{dfrac{3-sqrt{5}}{2-sqrt{3}}}
g)dfrac{sqrt{2+sqrt{3}}}{3+sqrt{3}}
h)sqrt{4-sqrt{15}}+sqrt{4+sqrt{15}}-2sqrt{3-sqrt{5}}
Đọc tiếp
Tính (rút gọn)
a) \(\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-}\sqrt{15}\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3+2}}\)
d) \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
f)\(\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\)
g)\(\dfrac{\sqrt{2+\sqrt{3}}}{3+\sqrt{3}}\)
h)\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
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\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
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\(d,\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{4+\sqrt{5}-1}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\sqrt{24+8\sqrt{5}}\left(\sqrt{5}-1\right)\\ =\sqrt{\left(2\sqrt{5}+2\right)^2}\left(\sqrt{5}-1\right)\\ =2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)\\ =8\)
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Xem thêm câu trả lời
Rút gọnsqrt{5+sqrt{21}}+sqrt{5-sqrt{21}}-2sqrt{4sqrt{7}}sqrt{8+sqrt{55}}-sqrt{8-sqrt{55}}-sqrt{125}left(sqrt{14}-sqrt{10}right)left(6-sqrt{35}right)sqrt{6+sqrt{35}}left(sqrt{2}+1right)left(sqrt{3}-1right)left(sqrt{6}+1right)left(5-2sqrt{2}-sqrt{3}right)sqrt{7-3sqrt{5}}left(7+3sqrt{5}right)left(3sqrt{2}+sqrt{10}right)
Đọc tiếp
Rút gọn
\(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}-2\sqrt{4\sqrt{7}}\)\(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)\(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\sqrt{6+\sqrt{35}}\)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)\(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
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cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
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Tính:
\(a)A=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\\ b)\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
a, Ta có : \(A=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
=> \(A=3-\sqrt{5}+\sqrt{5}-2=1\)
b, Ta có : \(B=\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=> \(B=\sqrt{2}\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=> \(B=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{6-2\sqrt{5}}\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{5-2\sqrt{5}+1}\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(5-1\right)\sqrt{36-20}=4.\sqrt{16}=4.4=16\)
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15 tháng 5 2021 lúc 21:55
Rút gọn:
\(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
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`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=\sqrt2(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)(\sqrt5-\sqrt3)`
`=(4+\sqrt{15})(8-2\sqrt{15})`
`=2(4+\sqrt{15})(4-\sqrt{15})`
`=2(16-15)`
`=2`
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a) \(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
\(=2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5}-1}\)
\(=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{5}+1\right)}{\sqrt{2}}\)
\(=2\cdot4=8\)
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Rút gọn
1) \(E=\left(\sqrt{11}-3\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{54}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
2) \(F=\frac{\left(\sqrt{3-\sqrt{5}}-1\right)\left(\sqrt{3-\sqrt{5}}\left(3-\sqrt{5}\right)+1\right)}{4-\sqrt{5}-\sqrt{3-\sqrt{5}}}+\sqrt{5}\)
rg: \(\sqrt{\left(\sqrt{7}-4\right)}^2\) = 3
chứng minh:
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\dfrac{1}{10}}+10\right)=3.3\sqrt{10}\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}\left(5\sqrt{\dfrac{1}{2}}+12\right)=-14.5\sqrt{2}\)
a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
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