Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)

=>\(m^2\ne1\)

=>\(m\notin\left\{1;-1\right\}\)

Khi \(m\notin\left\{1;-1\right\}\) thì \(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y-2m=0\end{matrix}\right.\)

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

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

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

Để \(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=2\\\dfrac{m}{m+1}>=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2\left(m+1\right)}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{-1}{m+1}>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{m+1}>=0\\-\dfrac{1}{m+1}>=0\end{matrix}\right.\Leftrightarrow m+1< 0\)

=>m<-1

Lời giải:
Từ PT$(1)\Rightarrow x=m+1-my$. Thay vô PT(2):

$m(m+1-my)+y=3m-1$

$\Leftrightarrow y(1-m^2)+m^2+m=3m-1$

$\Leftrightarrow y(1-m^2)=-m^2+2m-1(*)$

Để hpt có nghiệm $(x,y)$ duy nhất thì pt $(*)$ cũng phải có nghiệm $y$ duy nhất 

Điều này xảy ra khi $1-m^2\neq 0\Leftrightarrow m\neq \pm 1$
Khi đó: $y=\frac{-m^2+2m-1}{1-m^2}=\frac{-(m-1)^2}{-(m-1)(m+1)}=\frac{m-1}{m+1}$

$x=m+1-my=m+1-\frac{m(m-1)}{m+1}=\frac{3m+1}{m+1}$

Có:

$x+y=\frac{m-1}{m+1}+\frac{3m+1}{m+1}=\frac{4m}{m+1}<0$

$\Leftrightarrow -1< m< 0$

Kết hợp với đk $m\neq \pm 1$ suy ra $-1< m< 0$ thì thỏa đề.

a. Với `m=1`, ta có HPT: \(\left\{{}\begin{matrix}x+2y=18\\x-y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-6\\3y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=8\end{matrix}\right.\)

b. Theo đề bài `=>` \(\left\{{}\begin{matrix}mx+2y=18\\x-y=-6\\2x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx+2y=18\\x=1\\y=7\end{matrix}\right.\)

`=> m=4`

a) Thay m=1 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x+2y=18\\x-y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=24\\x-y=-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=8\\x=-6+y=-6+8=2\end{matrix}\right.\)

a) Thay m=-1 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}3x+y=7\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

Vậy: Khi m=-1 thì (x,y)=(1;4)

b) Ta có: \(\left\{{}\begin{matrix}3x+y=2m+9\\x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+y=2m+9\\x=5-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left(5-y\right)+y=2m+9\\x=5-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15-3y+y=2m+9\\x=5-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2y=2m-6\\x=5-y\end{matrix}\right.\)

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

Ta có: \(x^2+2y^2=18\)

\(\Leftrightarrow\left(m+2\right)^2+2\cdot\left(-m+3\right)^2=18\)

\(\Leftrightarrow m^2+4m+4+2\left(m^2-6m+9\right)-18=0\)

\(\Leftrightarrow m^2+4m-14+2m^2-12m+18=0\)

\(\Leftrightarrow3m^2-8m+4=0\)

\(\Leftrightarrow3m^2-2m-6m+4=0\)

\(\Leftrightarrow m\left(3m-2\right)-2\left(3m-2\right)=0\)

\(\Leftrightarrow\left(3m-2\right)\left(m-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3m-2=0\\m-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3m=2\\m=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{2}{3}\\m=2\end{matrix}\right.\)

Lời giải:

a.

Từ $x+y=2\Rightarrow y=2-x$. Thay vào PT(2):
$(m+1)x+m(2-x)=7$

$\Leftrightarrow x+2m=7$

$\Leftrightarrow x=7-2m$

$y=2-x=2-(7-2m)=2m-5$

Vậy hpt có nghiệm $(x,y)=(7-2m, 2m-5)(*)$

Nếu $x,y$ có 1 số $\geq 0$, một số $\leq 0$ thì $xy\leq 0< 1$

Nếu $x,y$ cùng $\geq 0$ thì áp dụng BĐT Cô-si:

$2=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq 1$

Vậy tóm lại $xy\leq 1(**)$
Từ $(*); (**)$ suy ra với mọi $m$ thì hpt luôn có nghiệm $x,y$ thỏa mãn $xy\leq 1$

b.

$xy>0$

$\Leftrightarrow (7-2m)(2m-5)>0$

$\Leftrightarrow 7> 2m> 5$

$\Leftrightarrow \frac{7}{2}> m> \frac{5}{2}$

Do $m$ nguyên nên $m=3$

Thử lại thấy đúng.

Linh tinh đếyyy ạ. Có gì sai thông cảm nhaaaa

a. Thay m = 1 ta được 

\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

b, Để hpt có nghiệm duy nhất khi \(\dfrac{1}{2}\ne-\dfrac{2}{3}\)*luôn đúng*

\(\left\{{}\begin{matrix}2x+4y=2m+6\\2x-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=m+6\\x=m+3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m+6}{7}\\x=m+3-2\dfrac{m+6}{7}\end{matrix}\right.\)

\(\Leftrightarrow x=m+3-\dfrac{2m+12}{7}=\dfrac{7m+21-2m-12}{7}=\dfrac{5m+9}{7}\)

Ta có : \(\dfrac{m+6}{7}+\dfrac{5m+9}{7}=-3\Rightarrow6m+15=-21\Leftrightarrow m=-6\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\2\left(4-2y\right)-3y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4-2y\\8-4y-3y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4-2y\\7y=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\rightarrow\left(x,y\right)=\left(2,1\right)\)

\(b,\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m+6\left(1\right)\\2x-3y=m\left(2\right)\end{matrix}\right.\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}7y=m+6\\x+2y=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+9}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\Rightarrow\) HPT có n_o duy nhất 

\(\left(x,y\right)=\left(\dfrac{5m+9}{7};\dfrac{m+6}{7}\right)\)

\(x+y=-3\)

\(\dfrac{5m+9}{7}+\dfrac{m+6}{7}=-3\)

\(\Leftrightarrow5m+9+m+6=-21\)

\(\Leftrightarrow6m=-36\Rightarrow m=-6\)

Với m = -6 thì hệ pt có n_o duy nhất TM x + y = -3