Question

[minh hanh dao]

cho hệ phương trình{ x+my= m+1

                                   mx+y=3m-1

tìm m để hpt No duy nhất mà x=|y|

Answer 1

[Trương Huy Hoàng]

\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=3m-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y=3m-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-my\\y\left(m^2-1\right)=m^2-2m+1\end{matrix}\right.\)

Với m = 1 ta có: \(\left\{{}\begin{matrix}x=2-y\\0y=0\left(VSN\right)\end{matrix}\right.\)

\(\Rightarrow\) Hpt vô số nghiệm

Với m = -1 ta có: \(\left\{{}\begin{matrix}x=y\\0y=4\left(VN\right)\end{matrix}\right.\)

\(\Rightarrow\) Hpt vô nghiệm

Với m \(\ne\) \(\pm\)1 ta có: \(\left\{{}\begin{matrix}x=m+1-my\\y=\dfrac{m^2-2m+1}{m^2-1}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m+1-\dfrac{m\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=m+1-\dfrac{m\left(m-1\right)}{m+1}=m+1-\dfrac{m^2-m}{m+1}\\y=\dfrac{m^2-2m+1}{m^2-1}=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=\dfrac{m-1}{m+1}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\\y=\dfrac{m-1}{m+1}\end{matrix}\right.\)

Vậy hpt có nghiệm duy nhất x = ..; y = ... với x \(\ne\) \(\pm\) 1

Ta có: x = |y|

\(\Leftrightarrow\) \(\dfrac{3m+1}{m+1}=\left|\dfrac{m-1}{m+1}\right|\) 

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\dfrac{3m+1}{m+1}=\dfrac{m-1}{m+1}\\\dfrac{3m+1}{m+1}=\dfrac{1-m}{m+1}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}3m+1=m-1\\3m+1=1-m\end{matrix}\right.\) (Vì m \(\ne\) -1)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2m=-2\\4m=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=-1\\m=0\end{matrix}\right.\) 

Vì m \(\ne\) -1 nên m = -1 KTM

\(\Rightarrow\) m = 0 thỏa mãn đk

Vậy m = 0

Chúc bn học tốt!