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a/ - Với \(x>\frac{1}{4}\) PT vô nghiêm
- Với \(x\le\frac{1}{4}\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(1-4x\right)^2\)
\(\Leftrightarrow\left(x^2+4x-2\right)\left(x^2-4x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+4x-2=0\\x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{6}\left(l\right)\\x=-2-\sqrt{6}\\x=4\left(l\right)\\x=0\end{matrix}\right.\)
2.
- Với \(x\ge-\frac{1}{4}\Leftrightarrow4x+1=x^2+2x-4\)
\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)
- Với \(x< -\frac{1}{4}\)
\(\Leftrightarrow-4x-1=x^2+2x-4\)
\(\Leftrightarrow x^2+6x-3=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)
3.
- Với \(x\ge\frac{5}{3}\)
\(\Leftrightarrow3x-5=2x^2+x-3\)
\(\Leftrightarrow2x^2-2x+2=0\left(vn\right)\)
- Với \(x< \frac{5}{3}\)
\(\Leftrightarrow5-3x=2x^2+x-3\)
\(\Leftrightarrow2x^2+4x-8=0\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)
4. Do hai vế của pt đều không âm, bình phương 2 vế:
\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)
\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}4x+3>=0\\\left(x+2-4x-3\right)\left(x+2+4x+3\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(-3x-1\right)\left(5x+5\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(3x+1\right)\left(x+1\right)>0\end{matrix}\right.\)
\(\Leftrightarrow x>-\dfrac{1}{3}\)
d: \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2< 0\\2x+1>=0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-2>=0\\\left(2x+1-3x+2\right)\left(2x+1+3x-2\right)>=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{2}{3}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(-x+3\right)\left(5x-1\right)>=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}< x< \dfrac{2}{3}\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-3\right)\left(5x-1\right)< =0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{2}< x< \dfrac{2}{3}\\\dfrac{2}{3}< =x< =3\end{matrix}\right.\)
\(\left(2x+1\right)\left(x-1\right)>0\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -\frac{1}{2}\end{matrix}\right.\)
\(\left(3x+1\right)\left(x-5\right)\left(-4x+5\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{3}\\\frac{5}{4}\le x\le5\end{matrix}\right.\)
\(\frac{x+2}{x-2}\le\frac{3x+1}{2x-1}\Leftrightarrow\frac{3x+1}{2x-1}-\frac{x+2}{x-2}\ge0\)
\(\Leftrightarrow\frac{x^2-8x}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{x\left(x-8\right)}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le0\\\frac{1}{2}< x< 2\\x\ge8\end{matrix}\right.\)