bài III:
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\y\ne-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x-3+3}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x-2+2}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{2}{y+2}=4-3=1\\\dfrac{2}{x-1}+\dfrac{1}{y+2}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{2}{y+2}=1\\\dfrac{4}{x-1}+\dfrac{2}{y+2}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{x-1}=7\\\dfrac{2}{x-1}+\dfrac{1}{y+2}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=1\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\left(nhận\right)\)
2:
a: Phương trình hoành độ giao điểm là:
\(x^2=3x+m^2-1\)
=>\(x^2-3x-m^2+1=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-m^2+1\right)=9+4m^2-4=4m^2+5>0\forall m\)
=>(P) luôn cắt (d) tại hai điểm phân biệt
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=-m^2+1\end{matrix}\right.\)
\(\left(x_1+1\right)\left(x_2+1\right)=1\)
=>\(x_1x_2+\left(x_1+x_2\right)+1=1\)
=>\(-m^2+1+3=0\)
=>\(m^2=4\)
=>\(\left[{}\begin{matrix}m=2\\m=-2\end{matrix}\right.\)
