1: Tọa độ giao điểm của (d) và (d') là:
\(\left\{{}\begin{matrix}x+m+1=-x+3m-1\\y=x+m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=3m-1-m-1=2m-2\\y=x+m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m-1\\y=m-1+m+1=2m\end{matrix}\right.\)
Thay x=m-1 và y=2m vào y=3x-1, ta được:
3(m-1)-1=2m
=>3m-4=2m
=>m=4
2: \(\text{Δ}=\left(-n\right)^2-4\left(n-1\right)\)
\(=n^2-4n+4=\left(n-2\right)^2>=0\forall n\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>(n-2)2>0
=>\(n-2\ne0\)
=>\(n\ne2\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=n\\x_1x_2=\dfrac{c}{a}=n-1\end{matrix}\right.\)
\(\dfrac{1}{\left(x_1-2\right)^2}+\dfrac{1}{\left(x_2-2\right)^2}=2\)
=>\(\dfrac{\left(x_1-2\right)^2+\left(x_2-2\right)^2}{\left[\left(x_1-2\right)\left(x_2-2\right)\right]^2}=2\)
=>\(\dfrac{x_1^2+x_2^2-4\left(x_1+x_2\right)+8}{\left[x_1x_2-2\left(x_1+x_2\right)+4\right]^2}=2\)
=>\(\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-4\left(x_1+x_2\right)+8}{\left[n-1-2n+4\right]^2}=2\)
=>\(\dfrac{n^2-2\left(n-1\right)-4n+8}{\left(-n+3\right)^2}=2\)
=>\(n^2-2n+2-4n+8=2\left(n-3\right)^2\)
=>\(2\left(n^2-6n+9\right)=n^2-6n+10\)
=>\(2n^2-12n+18-n^2+6n-10=0\)
=>\(n^2-6n+8=0\)
=>(n-2)(n-4)=0
=>\(\left[{}\begin{matrix}n=2\left(loại\right)\\n=4\left(nhận\right)\end{matrix}\right.\)
