a) Có lẽ đề có vấn đề.

b) \(\frac{x-11}{y-10}=\frac{11}{10}\Rightarrow10\left(x-11\right)=11\left(y-10\right)\)

\(10x-110=11y-110\)

\(10x-11y-110+110=0\)

\(10x-11y=0\)

\(10x-\left(10y+y\right)=0\)

\(10x-10y-y=0\)

\(10\left(x-y\right)-y=0\)

TH1: x-y = -12

10 (-12) -y =0

-120 - y =0

y = -120

Thay y = -120 vào x-y = -12

x - (-120) = -12

x + 120 = -12

x= -12 - 120

x= -132

TH2: x-y = 12

10 * 12 -y = 0

120 - y =0

y = 120

Thay y= 120 vào x-y = 12 

x - 120 = 12

x= 12 + 120

x= 132

Vậy nếu  y= -120 thì x= -132

nếu y= 120 thì x= 132

a) \(xy+3x-2y=11\)

\(\Leftrightarrow x\left(y+3\right)-2y-6=5\)

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

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

Lập bảng mà tìm 

b) \(x+y=xy\)

\(\Leftrightarrow x+y-xy=0\)

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

\(\Leftrightarrow x-1+y\left(x-1\right)=-1\)

\(\Leftrightarrow\left(x-1\right)\left(1+y\right)=-1\)

Tìm nốt

nhiều lắm bn, lm sao tìm đc hết

a, xy+3x-7y=21

=>xy+3x-7y-21=0

=>x(y+3)-7(y+3)=0

=>(x-7)(y+3)=0

=>x=7,y=-3

b,xy+3x-2y=11

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

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

=>x-2,y+3 thuộc ước của 5 cứ thế mà làm

CHÚC BẠN HỌC TỐT

\(a)xy+3x-7y=21\)

\(\Leftrightarrow xy+3x-7y-21=0\)

\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)=0\)

\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)=0\)

\(\Leftrightarrow\left(y+3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y+3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=-3\\x=7\end{cases}}}\)

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

\(\Leftrightarrow\left(xy+3x\right)-2y=6+5\)

\(\Leftrightarrow x\left(y+3\right)-2y-6=5\)

\(\Leftrightarrow x\left(y+3\right)-\left(2y+6\right)=5\)

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

\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\Rightarrow\left(y+3\right);\left(x-2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Xét các trường hợp :

• \(\hept{\begin{cases}y+3=5\\x-2=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}}\)
• \(\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)
• \(\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)
• \(\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)

b: \(\Leftrightarrow\left(x-3;y+2\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(4;9\right);\left(14;-1\right);\left(2;-13\right);\left(-8;-3\right)\right\}\)

c: \(\Leftrightarrow x\left(y-3\right)-y+3=3\)

=>(y-3)(x-1)=3

\(\Leftrightarrow\left(x-1;y-3\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)