1: ĐKXĐ: x<>3
\(\left\{{}\begin{matrix}\dfrac{2}{x-3}-3y=1\\\dfrac{3}{x-3}+2y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x-3}-6y=2\\\dfrac{9}{x-3}+4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{x-3}=26\\\dfrac{2}{x-3}-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-3=\dfrac{13}{26}=\dfrac{1}{2}\\3y=\dfrac{2}{x-3}-1=2:\dfrac{1}{2}-1=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{2}\\y=1\end{matrix}\right.\left(nhận\right)\)
2:
a: Phương trình hoành độ giao điểm là:
\(x^2=\left(m+2\right)x-m\)
=>\(x^2-\left(m+2\right)x+m=0\)
\(\text{Δ}=\left[-\left(m+2\right)\right]^2-4m=m^2+4>0\forall m\)
=>(P) luôn cắt (d) tại hai điểm phân biệt
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+2\\x_1x_2=\dfrac{c}{a}=m\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{x_1+x_2-2}\)
=>\(\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{x_1+x_2-2}\)
=>\(\dfrac{m+2}{m}=\dfrac{1}{m+2-2}=\dfrac{1}{m}\)
=>m+2=1
=>m=-1(nhận)
