b) \(\left|2x-\dfrac{1}{5}\right|\left(2x^2+3\right)=0\)
\(\left|2x-\dfrac{1}{5}\right|=0\) hoặc \(2x^2+3=0\)
*) \(\left|2x-\dfrac{1}{5}\right|=0\)
\(2x-\dfrac{1}{5}=0\)
\(2x=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}:2\)
\(x=\dfrac{1}{10}\)
*) \(2x^2+3=0\)
\(2x^2=-3\)
\(x^2=-\dfrac{3}{2}\) (vô lý)
Vậy \(x=\dfrac{1}{10}\)
d) \(\left|1-\dfrac{3}{2}\left|x-3\right|\right|.\sqrt{x+2}=0\)
\(\left|1-\dfrac{3}{2}\left|x-3\right|\right|.\sqrt{x+2}=0\)
\(\left|1-\dfrac{3}{2}\left|x-3\right|\right|=0\) hoặc \(\sqrt{x+2}=0\)
*) \(\left|1-\dfrac{3}{2}\left|x-3\right|\right|=0\)
\(1-\dfrac{3}{2}\left|x-3\right|=0\)
\(\dfrac{3}{2}\left|x-3\right|=1\)
\(\left|x-3\right|=1:\dfrac{3}{2}\)
\(\left|x-3\right|=\dfrac{2}{3}\)
\(x-3=-\dfrac{2}{3}\) hoặc \(x-3=\dfrac{2}{3}\)
**) \(x-3=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+3\)
\(x=\dfrac{7}{3}\)
**) \(x-3=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+3\)
\(x=\dfrac{11}{3}\)
*) Do \(x\ge0\Rightarrow x+2>0\)
\(\Rightarrow\sqrt{x+2}=0\) là vô lý
Vậy \(x=\dfrac{7}{3};x=\dfrac{11}{3}\)
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