( x- 7 ) . ( y + 3 ) = ? 

Đề thiếu rồi bạn ơi 

thank nha mk viết thiếu

\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)

=> x+3 = y+5 

=> x -y =5-3

=> x -y =2

\(y\left(x+4\right)-12x\left(2-y\right)-13xy+24x=2004\)

\(\Leftrightarrow xy+4y-24x+12xy-13xy+24x=2004\)

\(\Leftrightarrow4y=2004\Leftrightarrow y=501\)

=> Biểu thức trên không phụ thuộc vào giá trị biến x

\(\Rightarrow\left\{{}\begin{matrix}x\in R\\y=501\end{matrix}\right.\)

\(xy+4y-24x+12xy-13xy+24x-2004=0\)

\(4y-2004=0\)

\(y=501\)

\(y\left(x+4\right)-12x\left(2-y\right)-13xy+24x=2004\\ \Leftrightarrow xy+4y-24x+12xy-13xy+24x=2004\\ \Leftrightarrow4y=2004\Leftrightarrow y=501\)

Thay vào bt ban đầu

\(\Leftrightarrow501\left(x+4\right)-12x\left(2-501\right)-13x\cdot501+24x=2004\\ \Leftrightarrow501x+2004-12x\cdot\left(-499\right)-6513x+24x=2004\\ \Leftrightarrow0x+2004=2004\\ \Leftrightarrow x=0\)

Vậy \(\left(x;y\right)=\left(0;501\right)\)

xy+4y-24x+12xy-13xy+24x=2004

(xy+12xy-13xy)+(-24x+24x)+4y=2004

4y=2004

y=501

\(y\left(x+4\right)-12x\left(2-y\right)-13xy+24x=2004\)

\(\Leftrightarrow xy+4y-24x+12xy-13xy+24x=2004\)

\(\Leftrightarrow4y=2004\Leftrightarrow y=501\)

Vậy \(\left\{{}\begin{matrix}x\in R\\y=501\end{matrix}\right.\)

\(x\left(y-1\right)-y\left(y-1\right)\)

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

_________________

\(5x\left(x-2\right)-x+2\)

\(=5x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-1\right)\)

_________________

\(5x\left(x+1\right)-8x-8\)

\(=5x\left(x+1\right)-\left(8x+8\right)\)

\(=5x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(5x-8\right)\)