a) \(\left(2x^2+3x-6\right)^2-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\) \(\left(2x^2+3x-6-3x+2\right)\left(2x^2+3x-6+3x-2\right)=0\)
\(\Leftrightarrow\) \(\left(2x^2-4\right)\left(2x^2+6x-8\right)=0\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x^2-4=0\\2x^2+6x-8=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}2\left(x^2-2\right)=0\\2\left(x^2+3x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x^2-2=0\left(1\right)\\x^2+3x-4=0\left(2\right)\end{matrix}\right.\)
(1) \(x^2=2\) \(\Leftrightarrow x=\pm\sqrt{2}\)
(2) Vì \(a+b+c=1+3-4=0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt \(x_1=1\) ; \(x_2=\frac{c}{a}=-4\)
Vậy \(S=\left\{\pm\sqrt{2};1;-4\right\}\)
b) \(x^2-9x+20=0\)
\(\Delta=\left(-9\right)^2-4\times20=81-80=1\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt \(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-9\right)+\sqrt{1}}{2}=5\)
\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-9\right)-\sqrt{1}}{2}=4\)
Theo đề bài ta có hệ phương trình sau :
\(\left\{{}\begin{matrix}a+b=5\\ab=4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=5-b\\\left(5-b\right)b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\5b-b^2=4\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}a=5-b\\b^2-5b+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=5-1=4\\a=5-4=1\end{matrix}\right.\\\left[{}\begin{matrix}b=4\\b=1\end{matrix}\right.\end{matrix}\right.\) Vậy (a;b)=(4;1);(1;4)
