\(pt:x^2-4x+m=0\)
\(\Delta=\left(-4\right)^2-4.1.m=16-4m\)
Để pt có 2 nghiệm phân biệt thì \(16-4m>0\Leftrightarrow-4m>-16\Leftrightarrow m< 4\)
Theo hệ thức Vi-et:
\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m\end{matrix}\right.\)
Ta có: \(x^3_1+x_2^3-5\left(x^2_1+x^2_2\right)=26\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-5\left[\left(x_1+x_2\right)^2-2x_1x_2\right]=26\Leftrightarrow4^3-3.m.4-5\left[4-2m\right]=26\Leftrightarrow64-12m-20+10m=26\Leftrightarrow-2m=-18\Leftrightarrow m=9\left(KTM\right)\)
Vậy không có giá trị m thõa mãn
\(\Delta'=4-m\ge0\Leftrightarrow m\le4\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m\end{matrix}\right.\)
\(x_1^3+x_2^3-5\left(x_1^2+x_2^2\right)=26\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-5\left(x_1+x_2\right)^2+10x_1x_2=26\)
\(\Leftrightarrow64-12m-5.4^2+10m=26\)
\(\Leftrightarrow-2m=-18\Rightarrow m=9\)
