Bài 5:Thay x=1 và y=4 vào hệ phương trình, ta được:\(\left\{{}\begin{matrix}m+4n=17\\3\cdot n\cdot1+m\cdot4=-29\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m+4n=17\\4m+3n=-29\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}4m+16n=68\\4m+3n=-29\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13n=68+29=97\\m+4n=17\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}n=\dfrac{97}{13}\\m=17-4n=-\dfrac{167}{13}\end{matrix}\right.\)Bài 4:a: Thay m=2 vào hệ, ta được:\(\left\{{}\begin{matrix}x+y=1\\2x-y=4\cdot2=8\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x+y+2x-y=1+8\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}3x=9\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=3\\y=1-x=1-3=-2\end{matrix}\right.\)b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{1}{-1}\)=>\(m\ne-1\)c: Để hệ có nghiệm duy nhất thì m<>-1\(\left\{{}\begin{matrix}x+y=1\\mx-y=4m\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x+y+mx-y=1+4m\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x\left(m+1\right)=4m+1\\y=1-x\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=\dfrac{4m+1}{m+1}\\y=1-\dfrac{4m+1}{m+1}=\dfrac{m+1-4m-1}{m+1}=\dfrac{-3m}{m+1}\end{matrix}\right.\)x=3y=>\(\dfrac{4m+1}{m+1}=\dfrac{-9m}{m+1}\)=>-9m=4m+1=>-13m=1=>\(m=-\dfrac{1}{13}\left(nhận\right)\)Câu trả lời: Bài 5:Thay x=1 và y=4 vào hệ phương trình, ta được:\(\left\{{}\begin{matrix}m+4n=17\\3\cdot n\cdot1+m\cdot4=-29\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m+4n=17\\4m+3n=-29\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}4m+16n=68\\4m+3n=-29\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13n=68+29=97\\m+4n=17\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}n=\dfrac{97}{13}\\m=17-4n=-\dfrac{167}{13}\end{matrix}\right.\)Bài 4:a: Thay m=2 vào hệ, ta được:\(\left\{{}\begin{matrix}x+y=1\\2x-y=4\cdot2=8\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x+y+2x-y=1+8\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}3x=9\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=3\\y=1-x=1-3=-2\end{matrix}\right.\)b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{1}{-1}\)=>\(m\ne-1\)c: Để hệ có nghiệm duy nhất thì m<>-1\(\left\{{}\begin{matrix}x+y=1\\mx-y=4m\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x+y+mx-y=1+4m\\x+y=1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x\left(m+1\right)=4m+1\\y=1-x\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=\dfrac{4m+1}{m+1}\\y=1-\dfrac{4m+1}{m+1}=\dfrac{m+1-4m-1}{m+1}=\dfrac{-3m}{m+1}\end{matrix}\right.\)x=3y=>\(\dfrac{4m+1}{m+1}=\dfrac{-9m}{m+1}\)=>-9m=4m+1=>-13m=1=>\(m=-\dfrac{1}{13}\left(nhận\right)\)