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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
ĐKXĐ: \(x\ge\frac{2}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x+1}=a>0\\\sqrt{3x-2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x+3\)
Phương trình trở thành:
\(a-b=\frac{a^2-b^2}{5}\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-5\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{4x+1}=\sqrt{3x-2}\left(1\right)\\\sqrt{4x+1}+\sqrt{3x-2}=5\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x+1=3x-2\Rightarrow x=-3< \frac{2}{3}\left(l\right)\)
\(\left(2\right)\Leftrightarrow4x+1+3x-2+2\sqrt{\left(4x+1\right)\left(3x-2\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(4x+1\right)\left(3x-2\right)}=26-7x\) (\(\frac{2}{3}\le x\le\frac{26}{7}\))
\(\Leftrightarrow4\left(4x+1\right)\left(3x-2\right)=\left(26-7x\right)^2\)
\(\Leftrightarrow...\)