\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

=>4x+4y+y=6 và 3x+3y+y=8

=>4x+5y=6 và 3x+4y=8

=>12x+15y=18 và 12x+16y=32

=>-y=-14 và 4x+5y=6

=>y=14 và 4x=6-5y=6-70=-64

=>x=-16 và y=14

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)

b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\) 

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

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

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

a: =>2x-4+3+3y=-2 và 3x-6-2-2y=-3

=>2x+3y=-2+4-3=2-3=-1 và 3x-2y=-3+6+2=5

=>x=1; y=-1

b: =>x^2-x+xy-y=x^2+x-xy-y+2xy

=>-x-y=x-y và y^2+y-yx-x=y^2-2y+xy-2x-2xy

=>x=0 và y-x=-2y-2x

=>x=0 và y=0

=>xy-2x=xy-4x+2y-8 và 2xy+7x-6y-21=2xy+6x-7y-21

=>2x-2y=-8 và x+y=0

=>x-y=-4 và x+y=0

=>2x=-4 và x+y=0

=>x=-2 và y=2

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)