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a) \(2\sqrt{a^2}=2\left|a\right|=2a\) (vì \(a\ge0\))
b) \(\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) (vì \(a< 0\))
c) \(5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)
d) \(\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)=-\dfrac{1}{3}c^3\) (vì \(c< 0\Rightarrow c^3< 0\))
\(a)2\sqrt{a^2}=2.\left|a\right|=2a\) ( vì \(a\ge0\) )
\(b)\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) ( vì \(a< 0\) )
\(c)5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)
\(d)\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)\) ( vì \(c< 0\Rightarrow c^3< 0\) )
Chúc bn học tốt!
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
\(B=\dfrac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}}}{\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}}\right)\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}\right)\cdot2}-\sqrt{\left(6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}\right)\cdot2}}{2}\)
\(=\dfrac{\sqrt{12+4\sqrt{6}+4\sqrt{3}+4\sqrt{2}}-\sqrt{12-4\sqrt{6}+4\sqrt{3}-4\sqrt{2}}}{2}\)
\(=\dfrac{4}{2}\)
\(=2\)
\(C=\dfrac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{9-6\sqrt{2}}-\sqrt{6}\right)\sqrt{3}}{3}\)
\(=\dfrac{\sqrt{\left(9-6\sqrt{2}\right)\cdot3}-3\sqrt{2}}{3}\)
\(=\dfrac{\sqrt{27-18\sqrt{2}}-3\sqrt{2}}{3}\)
\(=\dfrac{\sqrt{\left(3-3\sqrt{2}\right)^2}-3\sqrt{2}}{3}\)
\(=\dfrac{3\sqrt{2}-3-3\sqrt{2}}{3}\)
\(=\dfrac{-3}{3}\)
\(=-1\)
3: |2x-1|=|x+1|
=>2x-1=x+1 hoặc 2x-1=-x-1
=>x=2 hoặc 3x=0
=>x=2 hoặc x=0
4: \(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{5}=0\\y-\sqrt{3}=0\\x-y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{5}\\y=\sqrt{3}\\z=x-y=-\sqrt{5}-\sqrt{3}\end{matrix}\right.\)
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
a; \(\sqrt{27a}\cdot\sqrt{3a}=\sqrt{81a^2}=9a\)
b: \(\dfrac{\sqrt{8a^4b^6}}{\sqrt{64a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=\dfrac{-\sqrt{2}}{4a}\)(do a<0)