b: Để d1 cắt d2 thì \(m\ne1\)
Phương trình hoành độ giao điểm là:
mx+3=x+1
=>mx-x=-2
=>x(m-1)=-2
=>\(x=-\dfrac{2}{m-1}\)
=>\(y=x+1=\dfrac{-2}{m-1}+1=\dfrac{-2+m-1}{m-1}=\dfrac{m-3}{m-1}\)
Để y=2x thì \(\dfrac{m-3}{m-1}=\dfrac{-4}{m-1}\)
=>m-3=-4
=>m=-1(nhận)
a: ĐKXĐ: x>=1 và \(y\ne-\dfrac{1}{2}x\)
\(\left\{{}\begin{matrix}2\sqrt{x-1}+\dfrac{3}{x+2y}=7\\3\sqrt{x-1}-\dfrac{1}{x+2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{x-1}+\dfrac{3}{x+2y}=7\\9\sqrt{x-1}-\dfrac{3}{x+2y}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x-1}=22\\2\sqrt{x-1}+\dfrac{3}{x+2y}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}=2\\\dfrac{3}{x+2y}=7-2\sqrt{x-1}=7-2\cdot2=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=4\\x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\2y=1-x=1-5=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=-2\end{matrix}\right.\left(nhận\right)\)
