Xét hệ phương trình :\(\hept{\begin{cases}mx-y=1\\\frac{x}{2}-\frac{y}{3}=334\end{cases}}\)

a, Khi m = 1 ta có hệ phương trình : \(\hept{\begin{cases}x-y=1\\3x-2y=2004\end{cases}\Leftrightarrow\hept{\begin{cases}x=2002\\y=2001\end{cases}}}\)

b, \(\hept{\begin{cases}mx-y=1\\\frac{x}{2}-\frac{y}{3}=334\end{cases}\Leftrightarrow\hept{\begin{cases}mx-y=1\\3x-2y=2004\end{cases}}}\)

Hệ phương trình vô nghiệm khi \(\frac{m}{3}=\frac{1}{2}\ne\frac{1}{2004}\Leftrightarrow m=\frac{3}{2}\)

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

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

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

\(P=\left(\dfrac{2m+3}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}=\left(2+\dfrac{1}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}\)

\(=4+\dfrac{4}{m+1}+\dfrac{4}{\left(m+1\right)^2}=\left(\dfrac{2}{m+1}+1\right)^2+3\ge3\)

\(P_{min}=3\) khi \(m=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=6m+4\\3x-2y=11-m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=6m+4\\5x=5m+15\end{matrix}\right.\)

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

b. \(P=\left(m+3\right)^2-\left(2m-1\right)^2\)

\(P=-3m^2+10m+10=-3\left(m-\dfrac{5}{3}\right)^2+\dfrac{55}{3}\le\dfrac{55}{3}\)

Dấu "=" xảy ra khi \(m=\dfrac{5}{3}\)

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từ phương trình thứ nhất ta có :

\(y=-x+3m+2\) thế xuống phương trình dười : \(3x+2x-6m-4=11-m\Leftrightarrow x=3+m\Rightarrow y=2m-1\)

b. ta có \(x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)

Dấu bằng xảy ra khi m=5/3

a, Khi \(m=-1\)ta có HPT : \(\hept{\begin{cases}-x+y=-2\\x-y=0\end{cases}}\)

=> HPT vô nghiệm

b, \(\hept{\begin{cases}mx+y=2m\\x+my=m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2m-mx\\x+m\left(2m-mx\right)=m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2m-mx\\\left(1-m^2\right)x=-2m^2+m+1\end{cases}}\)( * )

HPT vô nghiệm

<=> ( * ) vô nghiệm

\(\Leftrightarrow\hept{\begin{cases}1-m^2=0\\-2m^2+m+1\end{cases}}\ne0\)

<=> m = 1 hoặc m = -1 mà m khác 1 và -1/2 

<=> m = -1

ai tl ho vs @@