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H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
Y
30 tháng 5 2023
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
AT
4 tháng 7 2021
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
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) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)
\(=5\sqrt{10}+2\cdot5-5\sqrt{10}\)
\(=\left(5-5\right)\sqrt{10}+10\)
\(=0+10\)
\(=10\)
15 tháng 5 2021
`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`
15 tháng 5 2021
`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`
AT
7 tháng 7 2021
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
TG
7 tháng 7 2021
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
AT
7 tháng 7 2021
\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)
Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)
Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|\)
\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)
\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)
Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)
\(=2\sqrt{x-4}\)
Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
\(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
\(=\sqrt{\left(6+2\sqrt{5}\right)^2.\left(6+2\sqrt{5}\right)}+\sqrt{\left(6-2\sqrt{5}\right)^2.\left(6-2\sqrt{5}\right)}\)
\(=\left(6+2\sqrt{5}\right).\sqrt{\sqrt{5}^2+2\sqrt{5}+\sqrt{1}^2}+\left(6-2\sqrt{5}\right).\sqrt{\sqrt{5}^2-2\sqrt{5}+\sqrt{1}^2}\)
\(=\left(\sqrt{5}^2+2\sqrt{5}+\sqrt{1}^2\right).\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}+\left(\sqrt{5}^2-2\sqrt{5}+\sqrt{1}^2\right).\sqrt{\left(\sqrt{5}-\sqrt{1}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{1}\right)^2.\left(\sqrt{5}+\sqrt{1}\right)+\left(\sqrt{5}-\sqrt{1}\right)^2.\left(\sqrt{5}-\sqrt{1}\right)\)
\(=\left(\sqrt{5}+\sqrt{1}\right)^3+\left(\sqrt{5}-\sqrt{1}\right)^3\)
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