b) ĐKXĐ: \(x,y\neq 0\).

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\xy=-1\end{matrix}\right.\\2y=x^3+1\end{matrix}\right.\).

Với x - y = 0 suy ra x = y. Do đó \(2x=x^3+1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1=y\left(TMĐK\right)\\x=\pm\dfrac{\sqrt{5}-1}{2}=y\left(TMĐK\right)\end{matrix}\right.\).

Với xy = -1 suy ra \(y=-\dfrac{1}{x}\). Do đó \(x^3+\dfrac{2}{x}+1=0\Rightarrow x^4+x+2=0\). Phương trình vô nghiệm do \(x^4+x+2=\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\).

Vậy...

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

Ta có: \(\left\{{}\begin{matrix}3\left|x-1\right|+2\left(x-y\right)=4\\4\left|x-1\right|-\left(x-y\right)=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12\left|x-1\right|+8\left(x-y\right)=16\\12\left|x-1\right|-3\left(x-y\right)=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left(x-y\right)=-11\\3\left|x-1\right|+2\left(x-y\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\3\left|x-1\right|=4-2\left(x-y\right)=4-2\cdot\left(-1\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\left|x-1\right|=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y=-1\\x-1=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-y=-1\\x-1=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x+1=3+1=4\\x=3\end{matrix}\right.\\\left\{{}\begin{matrix}y=x+1=-1+1=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(3;4\right);\left(-1;0\right)\right\}\)

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

a: =>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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x-2\right)\left(y+1\right)=xy\\\left(x+8\right)\left(y-2\right)=xy\end{matrix}\right.\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\\\ \) \(\left\{{}\begin{matrix}xy+x-2y-2-xy=0\\xy-2x+8y-16-xy=0\end{matrix}\right.\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)\(\left\{{}\begin{matrix}x-2y=2\\-2x+8y=16\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-2y=2\\-x+4y=8\end{matrix}\right.\)\(\left\{{}\begin{matrix}2y=10\\x-2y=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}y=5\\x-10=2\end{matrix}\right.\)\(\left\{{}\begin{matrix}y=5\\x=12\end{matrix}\right.\)

Vậy hpt có nghiệm duy nhất là (x;y) = (12;5)

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

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x-2y-2-xy=0\\xy-2x+8y-16-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y-2=0\\-2x+8y-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=2\\-2x+8y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=4\\-2x+8y=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4y=20\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=2+2y=2+2\cdot5=12\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=12\\y=5\end{matrix}\right.\)

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

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

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

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

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(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) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

`{(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)}`