\(a,PT\Leftrightarrow-2x^2-10x+9=x^2+4x+4\\ \Leftrightarrow3x^2+14x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Thử lại ta thấy \(x=-5\) ko thỏa mãn
Vậy PT có nghiệm \(x=\dfrac{1}{3}\)
\(b,\Leftrightarrow\left[{}\begin{matrix}x^2-x-5=3x-5\left(x\le\dfrac{1-\sqrt{21}}{2};x\ge\dfrac{1+\sqrt{21}}{2}\right)\\x^2-x-5=5-3x\left(\dfrac{1-\sqrt{21}}{2}< x< \dfrac{1+\sqrt{21}}{2}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-4\right)=0\\x^2+2x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x\in\varnothing\end{matrix}\right.\\ \Leftrightarrow x=4\)
\(c,PT\Leftrightarrow3\left(\sqrt{x^2+5x+2}-4\right)=x^2+5x-14\\ \Leftrightarrow\dfrac{3\left(x^2+5x-14\right)}{\sqrt{x^2+5x+2}+4}-\left(x^2+5x-14\right)=0\\ \Leftrightarrow\left(x^2+5x-14\right)\left(\dfrac{3}{\sqrt{x^2+5x+2}+4}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)\left(x+7\right)=0\\\dfrac{3}{\sqrt{x^2+5x+2}+4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\\\sqrt{x^2+5x+2}=-1\left(vô.n_o\right)\end{matrix}\right.\)
Thử lại ta thấy \(x=2\left(tm\right);x=-7\left(tm\right)\)
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