\(\Leftrightarrow2xy+3x-5y-11=0\) nhân 2 vế với 2:

\(\Leftrightarrow4xy+6x-10y-22=0\\ \Leftrightarrow\left(4xy+6x\right)-\left(10y+15\right)-7=0\\ \Leftrightarrow\left(4xy+6x\right)-\left(10y+15\right)=7\\ \Leftrightarrow2x\left(2y+3\right)-5\left(2y+3\right)=7\\ \Leftrightarrow\left(2y+3\right)\left(2x-5\right)=7\)

TH1:\(\left\{{}\begin{matrix}2x-5=1\\2y+3=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

TH2:\(\left\{{}\begin{matrix}2x-5=7\\2y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-1\end{matrix}\right.\)

TH3:\(\left\{{}\begin{matrix}2x-5=-1\\2y+3=-7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)

TH4:\(\left\{{}\begin{matrix}2x-5=-7\\2y+3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

tìm giá trị nhỏ nhất của biểu thức (x^(2)-9)^(2)+|y-3|-1

\(5y-3x=2xy-11\)

\(\Rightarrow2xy+3x-5y-11=0\)

\(\Rightarrow4xy+6x-10y-22=0\)

\(\Rightarrow\left(4xy+6x\right)-\left(10y+15\right)=7\)

\(\Rightarrow2x\left(2y+3\right)-5\left(2y+3\right)=7\Rightarrow\left(2x-5\right)\left(2y+3\right)=7\)

Xét từng trường hợp :

1. \(\hept{\begin{cases}2x-5=1\\2y+3=7\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

2. \(\hept{\begin{cases}2x-5=7\\2y+3=1\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=-1\end{cases}}}\)

3. \(\hept{\begin{cases}2x-5=-1\\2y+3=-7\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-5\end{cases}}}\)

4. \(\hept{\begin{cases}2x-5=-7\\2y+3=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=-2\end{cases}}}\)

Vậy nghiệm của phương trình là : \(\left(x;y\right)=\left(-1;-2\right);\left(2;-5\right)\left(3;2\right);\left(6;-1\right)\)

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x=7:y=0

chắc  VẬY sai thì  thôi nhé

5y−3x=2xy−11 
\Rightarrow2xy+3x-5y-11=0⇒2xy+3x−5y−11=0 
\Rightarrow4xy+6x-10y-22=0⇒4xy+6x−10y−22=0 
\Rightarrow\left(4xy+6x\right)-\left(10y+15\right)=7⇒(4xy+6x)−(10y+15)=7 
\Rightarrow2x\left(2y+3\right)-5\left(2y+3\right)=7⇒2x(2y+3)−5(2y+3)=7 
\Rightarrow\left(2x-5\right)\left(2y+3\right)=7⇒(2x−5)(2y+3)=7 
Ta có các TH sau:
TH1: \left\{\begin{matrix}2x-5=1\\2y+3=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=3\\y=2\end{matrix}\right.{2x−5=12y+3=7​⇒{x=3y=2​ 
TH2: \left\{\begin{matrix}2x-5=-1\\2y+3=-7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\y=-5\end{matrix}\right.{2x−5=−12y+3=−7​⇒{x=2y=−5​ 
TH3: \left\{\begin{matrix}2x-5=7\\2y+3=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=6\\y=-1\end{matrix}\right.{2x−5=72y+3=1​⇒{x=6y=−1​ 
TH4: \left\{\begin{matrix}2x-5=-7\\2y+3=-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-1\\y=-2\end{matrix}\right.{2x−5=−72y+3=−1​⇒{x=−1y=−2​ 
Vậy......................................

5y - 3x = 2xy - 11

\(\Leftrightarrow\)2xy + 3x - 5y = 11

\(\Leftrightarrow\)4xy + 6x - 10y = 22

\(\Leftrightarrow\)2x(2y+3) - 5(2y+3) = 22-15

\(\Leftrightarrow\)(2x - 5)(2y+3)= 7

5y-3x=2xy-11

=>4xy+6x-10y=22

=>2x(2y+3)-5(2y+3)=7

=>(2x-5)(2y+3)=7

Đến đây dễ dàng hơn r

........................ !

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