Hệ có nghiệm duy nhất khi: \(\dfrac{3}{1}\ne\dfrac{m}{-2}\Rightarrow m\ne-6\)

Khi đó ta có:

\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+2my=10\\mx-2my=3m\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{-4}{m+6}\end{matrix}\right.\)

\(2x+y=1\Rightarrow\dfrac{2\left(3m+10\right)}{m+6}+\dfrac{-4}{m+6}=1\)

\(\Leftrightarrow\dfrac{6m+16}{m+6}=1\)

\(\Rightarrow6m+16=m+6\)

\(\Rightarrow m=-2\)

Kết hợp điều kiện đề bài và pt thứ 2 của hệ ta được:

\(\left\{{}\begin{matrix}x-y=-6\\2x+y=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=7\end{matrix}\right.\)

Thế vào pt đầu:

\(m.1+2.7=18\Rightarrow m=4\)

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

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

a)

Khi m = 1, ta có:

{ x+2y=1+3   

  2x-3y=1

=> { x+2y=4

        2x-3y=1

=> { 2x+4y=8

        2x-3y=1

=> { x+2y=4

        2x-3y-2x-4y=1-8

=> { x=4-2y

       -7y = -7

=> { x = 2

        y = 1

Vậy khi m = 1 thì hệ phương trình có cặp nghệm

(x; y) = (2;1)

a) Thay m=1 vào HPT ta có: 

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

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

⇔\(\left\{{}\begin{matrix}2x+4y=8\\7y=7\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y)= (2;1)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4-2y=4-2=2\end{matrix}\right.\)

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(2;1)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2\left(m+3-2y\right)-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2m+6-4y-3y-m=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y+m+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y=-m-6\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-\dfrac{2m+12}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7m+21-2m-12}{7}=\dfrac{5m+9}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\)

Để hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3 thì \(\dfrac{5m+9}{7}+\dfrac{m+6}{7}=3\)

\(\Leftrightarrow6m+15=21\)

\(\Leftrightarrow6m=6\)

hay m=1

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3

a/ Thay  \(m=1\) vào hpt ta có :

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

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

Vậy...

b/ Ta có :

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

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

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

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

Vì \(\dfrac{2}{3}\ne\dfrac{-1}{2}\)

nên hệ luôn có nghiệm duy nhất 

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

=>\(\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y+3x-2y=2m+5\\2x+y=m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{7}m+\dfrac{5}{7}\\y=m-2\left(\dfrac{2}{7}m+\dfrac{5}{7}\right)=\dfrac{3}{7}m-\dfrac{10}{7}\end{matrix}\right.\)

Vậy: \(M\left(\dfrac{2}{7}m+\dfrac{5}{7};\dfrac{3}{7}m-\dfrac{10}{7}\right)\)

Để M nằm hoàn toàn phía bên trái đường thẳng \(x=\sqrt{3}\) thì \(\dfrac{2}{7}m+\dfrac{5}{7}< \sqrt{3}\)

=>\(2m+5< 3\sqrt{7}\)

=>\(2m< 3\sqrt{7}-5\)

=>\(m< \dfrac{3\sqrt{7}-5}{2}\)

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:

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

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