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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{3x+1}-\sqrt{2x+3}=\dfrac{x-2}{4}\)
\(\Leftrightarrow\dfrac{x-2}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{x-2}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{1}{4}\left(1\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow\sqrt{3x+1}+\sqrt{2x+3}=4\)
\(\Leftrightarrow5x+4+2\sqrt{6x^2+11x+3}=16\)
\(\Leftrightarrow2\sqrt{6x^2+11x+3}=12-5x\) (\(x\le\dfrac{12}{5}\))
\(\Leftrightarrow4\left(6x^2+11x+3\right)=\left(12-5x\right)^2\)
\(\Leftrightarrow x^2-164x+132=0\Rightarrow\left[{}\begin{matrix}x=82-8\sqrt{103}\\x=82+8\sqrt{103}>\dfrac{12}{5}\left(loại\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow3x^2-3x+\left(x+1-\sqrt{3x+1}\right)+\left(x+2-\sqrt{5x+4}\right)=0\)
\(\Leftrightarrow3\left(x^2-x\right)+\dfrac{x^2-x}{x+1+\sqrt{3x+1}}+\dfrac{x^2-x}{x+2+\sqrt{5x+4}}=0\)
\(\Leftrightarrow\left(x^2-x\right)\left(3+\dfrac{1}{x+1+\sqrt{3x+1}}+\dfrac{1}{x+2+\sqrt{5x+4}}\right)=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow...\)
2.
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt[3]{2-8x^3}=b\end{matrix}\right.\)
Ta được hệ:
\(\left\{{}\begin{matrix}\left(2a-1\right)b=a\\a^3+b^3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2ab\\\left(a+b\right)^3-3ab\left(a+b\right)=2\end{matrix}\right.\)
\(\Rightarrow8\left(ab\right)^3-6\left(ab\right)^2=2\)
\(\Leftrightarrow\left(ab-1\right)\left[4\left(ab\right)^2+ab+1\right]=0\)
\(\Leftrightarrow ab=1\Rightarrow a+b=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\ab=1\end{matrix}\right.\) \(\Leftrightarrow a=b=1\)
\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\4x^2+4x+1-2x^2+3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\2x^2+7x=0\end{matrix}\right.\)
=>x=0