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\(\sqrt[]{\sqrt[]{3-\sqrt[]{1-\sqrt[]{21-12\sqrt[]{3}}}}}\)
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1 tháng 10 2017
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6 tháng 6 2021
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
14 tháng 7 2021
1) ĐKXĐ: \(x\ge5\)
2) ĐKXĐ: \(\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)
5) ĐKXĐ: \(\left[{}\begin{matrix}x\le2\\x\ge3\end{matrix}\right.\)
16 tháng 9 2021
\(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}=3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
19 tháng 5 2022
\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(=3-\sqrt{3}-2\sqrt{3}+3=6-3\sqrt{3}\)
2
19 tháng 5 2022
`A=\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}`
`A=\sqrt{(\sqrt{3})^2-2.\sqrt{3}.2+3^2}-\sqrt{3^2-2.3.2\sqrt{3}+(2\sqrt{3})^2}`
`A=\sqrt{(\sqrt{3}-3)^2}-\sqrt{(3-2\sqrt{3})^2}`
`A=|\sqrt{3}-3|-|3-2\sqrt{3}|`
`A=(3-\sqrt{3})-(2\sqrt{3}-3)`
`A=3-\sqrt{3}-2\sqrt{2}+3`
`A=6-3\sqrt{3}`
LA
1
HT
1
30 tháng 7 2018
\(\sqrt{\sqrt{3}-\sqrt{1-\sqrt{21-12\sqrt{3}}}}\)
\(=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(2\sqrt{3}\right)^2-2\cdot2\sqrt{3}\cdot3+3^2}}}\)
\(=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(2\sqrt{3}-3\right)^2}}}\)
\(=\sqrt{\sqrt{3}-\sqrt{1-2\sqrt{3}+3}}\)
\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}\)
\(=\sqrt{1}=1\)
4 tháng 9 2016
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
4 tháng 9 2016
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
30 tháng 7 2018
a) ta có : \(\sqrt{\sqrt{3}-\sqrt{1-\sqrt{21-12\sqrt{3}}}}=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(2\sqrt{3}-3\right)^2}}}\)
\(=\sqrt{\sqrt{3}-\sqrt{1-2\sqrt{3}+3}}=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)
b) bài này đề có sai 1 chút . mk sữa lại nha
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\left(5+3\sqrt{2}\right)^2}\)
\(=5+3\sqrt{2}\)
c) bài này đề có sai 1 chút . mk sữa lại nha
ta có : \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{3-2\sqrt{5}+3}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\sqrt{5}+1=1\)