Question

cho pt \(x^2-2mx+2m-1=0\)

 tìm m sao cho pt có 2 nghiệm thỏa mãn  \(x_1^2-5x_1x_2+x_2^2=25\)

Answer 1

\(x^2-2mx+2m-1=0\)

\(\Delta'=m^2-2m+1=\left(m-1\right)^2\ge0\forall m\)

⇒ Phương trình có hai nghiệm .

Theo viét \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m-1\end{matrix}\right.\)

Có : \(x_1^2-5x_1x_2+x^2_2=25\Leftrightarrow\left(x_1+x_2\right)^2-7x_1x_2=25\) \(\Leftrightarrow4m^2-14m+7=25\Leftrightarrow4m^2-14m-18=0\Leftrightarrow2m^2-7m-9=0\Leftrightarrow\left(2m-9\right)\left(m+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{9}{2}\\m=-1\end{matrix}\right.\)

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