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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+61x\ge0\\4x+2\ge0\\x^2+61x\le\left(4x+2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge0\\x\le-61\end{matrix}\right.\\x\ge-\dfrac{1}{2}\\15x^2-45x+4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge\dfrac{45+\sqrt{1785}}{30}\\x\le\dfrac{45-\sqrt{1785}}{30}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0\le x\le\dfrac{45-\sqrt{1785}}{30}\\x\ge\dfrac{45+\sqrt{1785}}{30}\end{matrix}\right.\)
\(ĐK:x\ge\dfrac{1}{3}\\ PT\Leftrightarrow-2x^2+14x-10+\left(4x-3\right)\left(x-2-\sqrt{3x-1}\right)=0\\ \Leftrightarrow-2\left(x^2-7x+5\right)+\dfrac{\left(4x-3\right)\left(x^2-7x+5\right)}{x-2+\sqrt{3x-1}}=0\\ \Leftrightarrow\left(x^2-7x+5\right)\left(\dfrac{4x-3}{x-2+\sqrt{3x-1}}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-7x+5=0\\\dfrac{4x-3}{x-2+\sqrt{3x-1}}=2\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow4x-3=2x-4+2\sqrt{3x-1}\\ \Leftrightarrow2x+1=2\sqrt{3x-1}\\ \Leftrightarrow4x^2+4x+1=12x-4\\ \Leftrightarrow4x^2-8x+5=0\left(\text{vô nghiệm}\right)\\ \Leftrightarrow x^2-7x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+\sqrt{29}}{2}\left(tm\right)\\x=\dfrac{7-\sqrt{29}}{2}\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(3-x\right)\left(3+x\right)}>0\)
Bảng xét dấu:
Từ bảng xét dấu ta có nghiệm của BPT là: \(x\in\left(-3;1\right)\cup\left(2;3\right)\)
\(\left|4x-1\right|=5-x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=5-x\left(x\ge\dfrac{1}{4}\right)\\4x-1=x-5\left(x< \dfrac{1}{4}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(nhận\right)\\x=-\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\)
2:
a: =>2x^2-4x-2=x^2-x-2
=>x^2-3x=0
=>x=0(loại) hoặc x=3
b: =>(x+1)(x+4)<0
=>-4<x<-1
d: =>x^2-2x-7=-x^2+6x-4
=>2x^2-8x-3=0
=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)