Question

Tìm các giá trị của tham số m sao cho phương trình \(x^2-2mx-4m-5=0\) có hai nghiệm\(x_1\),\(x_2\) thỏa mãn \(x_1^2\)-2(m-1)\(x_1\)+2\(x_2\)-4m=5+2\(x_1x_2\)

Answer 1

\(\Delta=\left(-2m\right)^2-4\cdot1\cdot\left(-4m-5\right)\)

\(=4m^2+16m+20\)

\(=4\left(m^2+4m+5\right)=4\left(m^2+4m+4+1\right)\)

\(=4\left(m+2\right)^2+4>=4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=-4m-5\end{matrix}\right.\)

\(x_1^2-2\left(m-1\right)x_1+2x_2-4m=5+2x_1x_2\)

=>\(x_1^2-2m\cdot x_1+2x_1+2x_2-4m=5+2x_1x_2\)

=>\(x_1^2-x_1\left(x_1+x_2\right)+2\left(x_1+x_2\right)-4m-5-2x_1x_2=0\)

=>\(-x_1x_2+2\left(x_1+x_2\right)-4m-5-2x_1x_2=0\)

=>\(2\left(x_1+x_2\right)-3x_1x_2-4m-5=0\)

=>\(2\cdot2m-3\cdot\left(-4m-5\right)-4m-5=0\)

=>\(4m+12m+15-4m-5=0\)

=>12m+10=0

=>12m=-10

=>\(m=-\dfrac{5}{6}\)