Để phương trình có 2 nghiệm phân biệt thì △>0\(\Leftrightarrow b^2-4ac>0\Leftrightarrow\left(-6\right)^2-4.1.\left(2m-3\right)>0\Leftrightarrow36-8m+12>0\Leftrightarrow8m< 48\Leftrightarrow m< 6\)
Theo định lí Vi-ét với m<6 ta có
\(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=\frac{6}{1}=6\\x_1x_2=\frac{c}{a}=\frac{2m-3}{1}=2m-3\end{matrix}\right.\)
Ta lại có \(\left(x_1^2-5x_1+2m-4\right)\left(x_2^2-5x_2+2m-4\right)=0\Leftrightarrow\left(x_1x_2\right)^2-5x_1^2x_2+\left(2m-4\right)x^2_1-5x_1x_2^2+25x_1x_2+5.\left(2m-4\right)x_1+\left(2m-4\right)x_2^2-5\left(2m-4\right)x_2+\left(2m-4\right)^2=0\Leftrightarrow\left(x_1x_2\right)^2-5x_1x_2\left(x_1+x_2\right)+\left(2m-4\right)\left(x_1^2+x_2^2\right)-5\left(2m-4\right)\left(x_1+x_2\right)+25x_1x_2+\left(2m-4\right)^2=0\Leftrightarrow\left(x_1x_2\right)^2-5x_1x_2\left(x_1+x_2\right)+\left(2m-4\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5\left(2m-4\right)\left(x_1+x_2\right)+25x_1x_2+\left(2m-4\right)^2=0\Leftrightarrow\left(2m-3\right)^2-5\left(2m-3\right).6+\left(2m-4\right)\left[36-2\left(2m-3\right)\right]-5\left(2m-4\right).6+25.\left(2m-3\right)+\left(2m-4\right)^2=0\Leftrightarrow4m^2-12m+9-60m+90+100m-8m^2-168-60m+120+50m-75+4m^2-16m+16=0\Leftrightarrow2m-8=0\Leftrightarrow m=4\left(tm\right)\)
Vậy m=4 thì phương trình trên có 2 nghiệm phân biệt thỏa mãn \(\left(x_1^2-5x_1+2m-4\right)\left(x_2^2-5x_2+2m-4\right)=0\)
\(\Delta'\ge0\Rightarrow m\le6\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=2m-3\end{matrix}\right.\)
Do \(x_1;x_2\) là nghiệm nên:
\(\left\{{}\begin{matrix}x_1^2-6x_1+2m-3=0\\x_2^2-6x_2+2m-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-5x_1+2m-4=x_1-1\\x_2^2-5x_2+2m-4=x_2-1\end{matrix}\right.\)
Thay vào bài toán:
\(\Leftrightarrow\left(x_1-1\right)\left(x_2-1\right)=0\Leftrightarrow x_1x_2-\left(x_1+x_2\right)+1=0\)
\(\Leftrightarrow2m-3-6+1=0\)
\(\Leftrightarrow2m=8\Rightarrow x=4\)
