a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)

$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$

`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$

`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$

Vậy HPT vô nghiệm

$\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|+2|y-1|=9\\2x+2|y-1|=-2\\\end{cases}$

`<=>` $\begin{cases}|x-2|-2x=11\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|=2x+11\\x+|y-1|=-1\\\end{cases}$

Đến đây dễ rồi bạn tự giải :D

`{(x+1)(y-1)=2),((x-3)(y+1)=-6):}`

`<=>{(xy-x+y-1=2),(xy+x-3y-3=-6):}`

`<=>{(x(y-1)=3-y),(xy+x-3y-3=-6):}`

`<=>{(x=[3-y]/[y-1]\text{   (1)}),(xy+x-3y=-3\text{    (2)}):}`

Thay `(1)` vào `(2)` có:

    `[3-y]/[y-1] .y+[3-y]/[y-1]-3y=-3`

  `=>3y-y^2+3-y-3y^2+3y=-3y+3`

`<=>4y^2-8y=0`

`<=>[(y=0),(y=2):}`

  `=>[(x=[3-0]/[0-1]=-3),(x=[3-2]/[2-1]=1):}`

Vậy `S={(-3;0),(1;2)}`

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

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 \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\) 

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

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

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

Giả sử \(x>y\Rightarrow x^3+x^2+x>y^3+y^2+y\)

\(\Rightarrow y^3>z^3\Leftrightarrow y>z\left(2\right)\)

\(\Rightarrow y^3+y^2+y>z^3+z^2+z\Rightarrow z>x\left(3\right)\)

Từ \(\left(2\right);\left(3\right)\Rightarrow y>x\) (Vô lí)

Giả sử \(x< y\Rightarrow x^3+x^2+x< y^3+y^2+y\)

\(\Rightarrow y^3< z^3\Leftrightarrow y< z\left(4\right)\)

\(\Rightarrow y^3+y^2+y< z^3+z^2+z\Rightarrow z< x\left(5\right)\)

Từ \(\left(4\right);\left(5\right)\Rightarrow y< x\) (Vô lí)

\(\Rightarrow x=y=z\)

\(\left(1\right)\Leftrightarrow x^3+x^2+x=x^3\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow x=y=z=0\) hoặc \(x=y=z=-1\)