a: Phương trình hoành độ giao điểm là:
\(-x^2=\left(3-2m\right)x-4\)
=>\(x^2+\left(3-2m\right)x-4=0\)
\(a\cdot c=1\cdot\left(-4\right)=-4< 0\)
=>(P) luôn cắt (d) tại hai điểm phân biệt
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m-3\\x_1x_2=\dfrac{c}{a}=-4\end{matrix}\right.\)
\(K=\left(1-x_1^2\right)\cdot\left(1-x_2^2\right)-2x_1-2x_2\)
\(=\left(x_1-1\right)\left(x_2-1\right)\left(x_1+1\right)\left(x_2+1\right)-2\left(x_1+x_2\right)\)
\(=\left[x_1x_2-\left(x_1+x_2\right)+1\right]\left[x_1x_2+\left(x_1+x_2\right)+1\right]-2\left(x_1+x_2\right)\)
\(=\left[-4-\left(2m-3\right)+1\right]\left[-4+\left(2m-3\right)+1\right]-2\left(2m-3\right)\)
\(=\left(-4-2m+3+1\right)\left(-4+2m-3+1\right)-4m+6\)
\(=\left(-2m\right)\left(2m-6\right)-4m+6\)
\(=-4m^2+12m-4m+6\)
\(=-4m^2+8m+6\)
\(=-4\left(m^2-2m-\dfrac{3}{2}\right)\)
\(=-4\left(m^2-2m+1-\dfrac{5}{2}\right)\)
\(=-4\left(m-1\right)^2+10< =10\forall m\)
Dấu '=' xảy ra khi m-1=0
=>m=1
