trừ 2 vế, rồi tìm y=>x

a) Thay m=3 vào hpt \(\hept{\begin{cases}x+y=1\\3x+2y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=1-x\\3x+2-2x=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}\)

Vậy m=3 thì hpt có nghiệm duy nhất (x,y)=(1;0)

b)Ta có  \(\hept{\begin{cases}x=1-y\\m-my+2y=m\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1-y\left(1\right)\\\left(2-m\right)y=0\left(2\right)\end{cases}}\)

Để hpt có nghiệm duy nhất \(\Leftrightarrow pt\left(2\right)\ne0\Leftrightarrow2-m\ne0\Leftrightarrow m\ne2\)

Khi đó \(\left(2\right)\Leftrightarrow y=0\).Thay vào \(\left(1\right)\Leftrightarrow x=1\)

Để hpt có vô số nghiệm \(\Leftrightarrow2-m=0\Leftrightarrow m=2\)

Vậy m\(\ne\)2 thì hpt có nghiệm duy nhất (x;y)=(1;0)

      m=2 thì hpt có vô số nghiệm

1) Thay \(m=\sqrt{3}+1\) vào hệ phương trình, ta được:

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

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

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-5+2\sqrt{3}}{13}\\3x=\sqrt{3}-\dfrac{12+10\sqrt{3}}{13}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-5+2\sqrt{3}}{13}\\x=\left(\dfrac{13\sqrt{3}-12-10\sqrt{3}}{13}\right)\cdot\dfrac{1}{3}=\dfrac{3\sqrt{3}-12}{13}\cdot\dfrac{1}{3}=\dfrac{\sqrt{3}-4}{13}\end{matrix}\right.\)

Vậy: Khi \(m=\sqrt{3}+1\) thì hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{3}-4}{13}\\y=\dfrac{-5+2\sqrt{3}}{13}\end{matrix}\right.\)

bạn ơi mk thử lại ko đúng

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+3y=1\\3x-3y=9\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}5x=10\\x-y=3\end{matrix}\right.\)

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

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

\(\left\{{}\begin{matrix}0,3x+0,5y=3\\1,5x-2y=1,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1,5x+2,5y=15\\1,5x-2y=1,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4,5y=13,5\\1,5x-2y=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\1,5x=2y+1,5=2\cdot3+1,5=7,5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}0,3x+0,5y=3\\1,5x-2y=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1,5x+2,5y=15\\1,5x-2y=1,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4,5y=-13,5\\1,5x-2y=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13,5}{4,5}=3\\1,5x-2.3=1,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1,5+6}{1,5}=5\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(5;3\right)\)