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\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{\sqrt{2}}{2}\right)\cdot3\sqrt{6}\\ =36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\\ =36-36\sqrt{2}+27\sqrt{3}\)
\(\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}\)
\(=12-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=12-6\sqrt{3.6}+10\sqrt{3.2^2}-\sqrt{3.4^2}\)
\(=12-6\sqrt{3}.\sqrt{6}+20\sqrt{3}-4\sqrt{3}\)
\(=12\sqrt{6}+\left(-6+20-4\right)\sqrt{3}\)
\(=12\sqrt{6}+10\sqrt{3}\)
mk ko biết mk làm có đúng ko nữa vì mk năm nay mới lên lớp 9 thoy
mk làm đc là do mk tự học
nếu thấy đúng thì k để mk biết nhé
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
a: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}+3\right)+2\sqrt{2}-3}{-1}\)
\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)
b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)
\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)
c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)
\(=\dfrac{-2\left(3-2\sqrt{2}\right)+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{-6+4\sqrt{2}+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}\)
\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)
\(\dfrac{2}{\sqrt[]{6}-2}+\dfrac{2}{\sqrt[]{6}+2}+\dfrac{5}{\sqrt[]{6}}\)
\(=\dfrac{2}{\sqrt[]{6}-2}+\dfrac{2}{\sqrt[]{6}+2}+\dfrac{5\sqrt[]{6}}{6}\)
\(=\dfrac{12\left(\sqrt[]{6}+2\right)}{6\left(\sqrt[]{6}-2\right)\left(\sqrt[]{6}+2\right)}+\dfrac{12\left(\sqrt[]{6}-2\right)}{6\left(\sqrt[]{6}-2\right)\left(\sqrt[]{6}+2\right)}+\dfrac{5\sqrt[]{6}\left(\sqrt[]{6}-2\right)\left(\sqrt[]{6}+2\right)}{6\left(\sqrt[]{6}-2\right)\left(\sqrt[]{6}+2\right)}\)
\(=\dfrac{12\sqrt[]{6}+24+12\sqrt[]{6}-24+5\sqrt[]{6}\left(6-2\right)}{6\left(6-2\right)}\)
\(=\dfrac{24\sqrt[]{6}+20\sqrt[]{6}}{24}\)
\(=\dfrac{44\sqrt[]{6}}{24}\)
\(=\dfrac{11\sqrt[]{6}}{6}\)
\(=\left[\sqrt{2.2.6}-\sqrt{4.4.3}+\sqrt{5.5.2}-\sqrt{\left(\frac{1}{4}\right)^2.8}\right].\sqrt{54}\)
\(=\left[\sqrt{24}-\sqrt{48}+\sqrt{50}-\sqrt{\frac{1}{2}}\right].\sqrt{54}\)
\(=\sqrt{24.54}-\sqrt{48.54}+\sqrt{50.54}-\sqrt{\frac{1}{2}.54}\)
\(=\sqrt{1296}-\sqrt{2592}+\sqrt{2700}-\sqrt{27}\)
\(=36-\sqrt{1296.2}+10\sqrt{27}-\sqrt{27}\)
\(=36-36\sqrt{2}+9\sqrt{27}\)
\(=36-36\sqrt{2}+27\sqrt{3}\)
1.
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
2.
a, ĐK: \(x\in R\)
\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\sqrt{2}+1\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
a) ( 75 - 3 2 - 12 )( 3 + 2 )
=(5 3 - 3 2 - 2 3 )( 3 + 2 )
=3( 3 - 2 )( 3 + 2 ) = 3
d) 2 3 - 6 8 - 2 = 12 - 6 2 2 - 2 = 6 2 - 1 2 2 - 1 = 6 2