\(a,2\left(x-1\right)-3x\left(x-5\right)=21\)
\(\Leftrightarrow2x-2-3x^2+15x-21=0\)
\(\Leftrightarrow-3x^2+17x-23=0\)
\(\Leftrightarrow-3\left(x^2+17x+\dfrac{298}{4}\right)+\dfrac{401}{2}=0\)
\(\Leftrightarrow-3\left(x+\dfrac{17}{2}\right)^2=-\dfrac{401}{2}\)
\(\Leftrightarrow\left(x+\dfrac{17}{2}\right)^2=\dfrac{401}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{17}{2}=\sqrt{\dfrac{401}{6}}\\x+\dfrac{17}{2}=-\sqrt{\dfrac{401}{6}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{401}{6}}-\dfrac{17}{2}\\x=-\sqrt{\dfrac{401}{6}}-\dfrac{17}{2}\end{matrix}\right.\)
\(b,\left(x+3\right)-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x+3-x^2-4x+32-1=0\)
\(\Leftrightarrow x^2-3x+34=0\)
\(\Leftrightarrow\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{127}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=-\dfrac{127}{4}\)
Ta có:
\(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
Vậy pt vô nghiệm
