\(\left\{{}\begin{matrix}y=5-mx\\2x-5+mx=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5-mx\\x\left(m+2\right)=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5-mx\\x=\dfrac{3}{m+2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5-m.\dfrac{3}{m+2}\\x=\dfrac{3}{m+2}\end{matrix}\right.\)

Ta co : x_o+y_o=1

=> 5-\(\dfrac{3m}{m+2}+\dfrac{3}{m+2}=1\)

=> \(\dfrac{5.\left(m+2\right)-3m+3}{m+2}=1\)

=> 5m+10-3m+3=m+2

=> 2m-m=2-13

=> m=-11

\(\left\{{}\begin{matrix}mx+y=5\left(1\right)\\2x-y=-2\left(2\right)\end{matrix}\right.\)

từ (1) ta có y=5-mx(3)

thế vào (2) ta có 2x-5+mx=-2\(\Leftrightarrow\) (2+m)x=3\(\Leftrightarrow\)x=\(\dfrac{3}{2+m}\)(4)

thế (4) vào (3) ta có

y=5-m\(\dfrac{3}{2+m}\)=\(\dfrac{10+2m}{2+m}\)

vậy hệ có nghiệm duy nhất là(\(\dfrac{3}{2+m}\);\(\dfrac{10+2m}{2+m}\))

mà x+y=1

\(\Rightarrow\)\(\dfrac{3}{2+m}+\dfrac{10+2m}{2+m}=1\)\(\Leftrightarrow\)m=-11

vậy m=-11

Để hệ có nghiệm duy nhất thì \(\frac{2}{-5}\ne\frac{3}{1}\)

\(\Leftrightarrow-15\ne2\) ( luôn đúng)

=> hệ luôn có nghiệm duy nhất

Với mọi m, hệ luôn có nghiệm duy nhất nên ta có:

\(\left\{{}\begin{matrix}2x+3y=m\\-5x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=m\\-15x+3y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}17x=m+3\\-5x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{m+3}{17}\\y=\frac{5m-2}{17}\end{matrix}\right.\)

Để x > 0, y>0 thì \(\left\{{}\begin{matrix}\frac{m+3}{17}>0\\\frac{5m-2}{17}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m+3>0\\5m-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-3\\m>\frac{2}{5}\end{matrix}\right.\)

Kết hợp 2 đk, ta được \(m>\frac{2}{5}\)

=.= hk tốt!!

Lấy pt 1 cộng vế với vế của pt 2 ta được

\(2x+y+x-y=m+2+m\Leftrightarrow3x=2m+2\Leftrightarrow x=\dfrac{2m+2}{3}\)

từ pt 2 ta suy ra \(y=\dfrac{-m+2}{3}\)

Để hpt có nghiệm \(x_0,y_0\) thoả mãn đk đề bài thì \(\dfrac{-m+2}{3}+\dfrac{2m+2}{3}=3\Leftrightarrow\dfrac{m+4}{3}=3\Leftrightarrow m=5\)

Vậy ..........

m=5

Để hệ có nghiệm duy nhất thì \(\frac12<>\frac{3}{-1}\)

=>0,5<>-3(đúng)

=>Hệ luôn có nghiệm duy nhất

\(\begin{cases}x-3y=5-2m\\ 2x+y=3\left(m+1\right)\end{cases}\)

=>\(\begin{cases}2x-6y=10-4m\\ 2x+y=3\left(m+1\right)\end{cases}\Rightarrow\begin{cases}2x+y-2x+6y=3\left(m+1\right)+4m-10\\ x-3y=5-2m\end{cases}\)

=>\(\begin{cases}7y=3m+3+4m-10\\ x=3y+5-2m\end{cases}\Rightarrow\begin{cases}y=m-1\\ x=3\left(m-1\right)+5-2m=3m-3+5-2m=m+2\end{cases}\)

\(x_0^2+y_0^2=9m\)

=>\(\left(m+2\right)^2+\left(m-1\right)^2=9m\)

=>\(m^2+4m+4+m^2-2m+1-9m=0\)

=>\(2m^2-7m+5=0\)

=>\(2m^2-2m-5m+5=0\)

=>(m-1)(2m-5)=0

=>m=1 hoặc m=5/2

\(\left\{{}\begin{matrix}x_0-my_0=2-4m\\mx_0+y_0=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0-2=m\left(y_0-4\right)\\y_0-1=m\left(3-x_0\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x_0-2\right)\left(3-x_0\right)=m\left(y_0-4\right)\left(3-x_0\right)\\\left(y_0-1\right)\left(y_0-4\right)=m\left(y_0-4\right)\left(3-x_0\right)\end{matrix}\right.\)

\(\Rightarrow\left(x_0-2\right)\left(3-x_0\right)=\left(y_0-1\right)\left(y_0-4\right)\)

\(\left\{{}\begin{matrix}mx+y=5\\2x-y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)x=3\\2x-y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{m+2}\\\frac{6}{m+2}-y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{m+2}\\y=\frac{10+2m}{m+2}\end{matrix}\right.\)

\(\Rightarrow x+y=\frac{3}{m+2}+\frac{10+2m}{m+2}=\frac{13+2m}{m+2}\)

\(\Leftrightarrow\frac{13+2m}{m+2}=1\Leftrightarrow13+2m=m+2\)

\(\Leftrightarrow m=-11\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-ay\\a\left(2-ay\right)-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-ay\\2a-a^2y-2y=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2a-1}{a^2+2}\\x=\dfrac{2a^2+4-2a^2+a}{a^2+2}=\dfrac{a+4}{a^2+2}\end{matrix}\right.\)

xy<0

=>(2a-1)*(a+4)/(a^2+2)^2<0

=>(2a-1)(a+4)<0

=>-4<a<1/2

mà a là số nguyên lớn nhất

nen a=0

\(\left\{{}\begin{matrix}x-2y=3-m\\2x+y=3m+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6m+12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3-m\\5x=5m+15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)

\(A=\left(m+3\right)^2+m^2=2m^2+6m+9=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(m+\dfrac{3}{2}=0\Rightarrow m=-\dfrac{3}{2}\)

mọi người ơi giúp mình vs mai ktra r