Câu 6:
a: Thay m=2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x+2y=1\\2x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=2\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-2y=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)(1)
\(\left\{{}\begin{matrix}x+my=1\\mx+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\m\left(1-my\right)+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\m-m^2y+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\m-y\left(m^2-1\right)=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1-my\\y\left(m^2-1\right)=m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m^2-1}=\dfrac{1}{m+1}\\x=1-\dfrac{m}{m+1}=\dfrac{m+1-m}{m+1}=\dfrac{1}{m+1}\end{matrix}\right.\)
x>0 và y>0
=>\(\left\{{}\begin{matrix}\dfrac{1}{m+1}>0\\\dfrac{1}{m+1}>0\end{matrix}\right.\)
=>m+1>0
=>m>-1
Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m>-1\\m\ne1\end{matrix}\right.\)
Bài 5:
a: Thay m=-2 vào hệ, ta được:
\(\left\{{}\begin{matrix}-2x-4y=-10\\3x+y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=5\\3x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=5\\6x+2y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-5x=5\\3x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3x=-3\cdot\left(-1\right)=3\end{matrix}\right.\)
Anh chị giúp em với ạ^^
