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Ta có: \(\sqrt{75}-2\cdot\sqrt{12}+\sqrt{3}\)
\(=5\sqrt{3}-2\cdot2\sqrt{3}+\sqrt{3}\)
\(=6\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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...
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x>=0 và x^2+x=x^2
=>x=0
b: =>x>=2 và x^2-4x-3=x^2-4x+4
=>-3=4(loại)
\(a)ĐK:x\ge0\)
\(pt\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy, pt có nghiệm duy nhất là x=0
\(b)ĐK:x\ge2+\sqrt{7}\)
\(pt\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4x-3=(x-2)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x-3=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3=4\end{matrix}\right.\)(vô lý)
Vậy pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=4\cdot2\sqrt{6}-3\cdot3\sqrt{6}+5\sqrt{6}-5\sqrt{6}\)
\(=8\sqrt{6}-9\sqrt{6}=-\sqrt{6}\)
b: \(=2\left(3-2\sqrt{2}\right)-2\left(3+2\sqrt{2}\right)\)
\(=6-4\sqrt{2}-6-4\sqrt{2}=-8\sqrt{2}\)
c: \(=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
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a: \(=5\cdot5\sqrt{3}-\dfrac{1}{3}\cdot3\sqrt{3}=24\sqrt{3}\)
b: \(=\dfrac{12\left(3+\sqrt{5}\right)}{4}=9+3\sqrt{5}\)
c: \(=3-\sqrt{5}+\sqrt{5}=3\)
a) ( 75 - 3 2 - 12 )( 3 + 2 )
=(5 3 - 3 2 - 2 3 )( 3 + 2 )
=3( 3 - 2 )( 3 + 2 ) = 3