Question

cho hso \(y=\dfrac{x^2-mx+m}{x^2+1}\). biết pt \(y'=0\) có 2 ng x1, x2. tìm m để \(x_1+x_2=3\)?

Answer 1

\(y'=\dfrac{\left(2x-m\right)\left(x^2+1\right)-2x\left(x^2-mx+m\right)}{\left(x^2+1\right)^2}=\dfrac{2x-mx^2-m+2mx^2-2mx}{\left(x^2+1\right)^2}=\dfrac{mx^2+2\left(1-m\right)x-m}{\left(x^2+1\right)^2}\)

\(y'=0\Leftrightarrow mx^2+2\left(1-m\right)x-m=0\)

Xet \(m=0\) ko thoa man pt

Xet \(m\ne0\)

\(\left\{{}\begin{matrix}\Delta'>0\\\dfrac{2\left(m-1\right)}{m}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(1-m\right)^2+m^2>0\left(ld\right)\\m=-2\end{matrix}\right.\Rightarrow m=-2\)