Giải giúp mk đi các bạn, mk cần gấp lắm
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a, \(2\sqrt{3}-\sqrt{4+x^2}=0\Leftrightarrow\sqrt{4+x^2}=2\sqrt{3}\)
\(\Leftrightarrow x^2+4=12\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)
b, \(\sqrt{16x+16}-\sqrt{9x+9}=0\)ĐK : x >= -1
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=0\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c, \(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow2\left|x+2\right|=8\Leftrightarrow\left|x+2\right|=4\)
TH1 : \(x+2=4\Leftrightarrow x=2\)
TH2 : \(x+2=-4\Leftrightarrow x=-6\)
c: Ta có: \(\sqrt{4\left(x+2\right)^2}=8\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
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4.2x-3=125
4.2x=125+3
4.2x=128
2x=128:4
2x=32
2x=25
-> x=5
vậy x=5
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\(\dfrac{1}{2022}\) \(\times\) \(\dfrac{2}{5}\) + \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{7}{5}\) - \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{8}{10}\)
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{2}{5}\) + \(\dfrac{7}{5}\) - \(\dfrac{8}{10}\))
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{9}{5}\) - \(\dfrac{4}{5}\))
= \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{5}{5}\)
= \(\dfrac{1}{2022}\times1\)
= \(\dfrac{1}{2022}\)
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a) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) \(\sqrt{\left(1-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+3-\sqrt{5}\)
=2
c) \(\sqrt{21-12\sqrt{3}}-\sqrt{13-4\sqrt{3}}\)
\(=2\sqrt{3}-3-2\sqrt{3}+1\)
=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
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