a) Khi m = 1, pt trở thành:
\(x^2-2x-15=0\\ \Leftrightarrow x^2+3x-5x-15=0\\ \Leftrightarrow x\left(x+3\right)-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(b)\Delta'=b'^2-ac\\ =\left(-m\right)^2-1\left(-4m-11\right)\\ =m^2+4m+11\\ =\left(m^2+2.m.2+2^2\right)+7\\ =\left(m+2\right)^2+7>\forall m\)
\(c)\)Theo hệ thức Vi - ét: \(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=2m\\x_1.x_2=\frac{c}{a}=-4m-11\end{matrix}\right.\)
\(\frac{x_1}{x_2-1}+\frac{x_2}{x_1-1}=-5\\ \Leftrightarrow\frac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}=-5\\ \Leftrightarrow\frac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}=-5\\ \Leftrightarrow\frac{\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}=-5\\ \Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}=-5\)
Thay vào là được nhé! Tự tiếp giúp mình
Nguyễn Việt Lâm giúp mk nhá, thanks bn nhìu :>>>
