a, \(4y^2+1-4y=\left(2y\right)^2-2.2y.1+1^2=\left(2y-1\right)^2\)

b, \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

c, \(x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

a. \(3xy+x+15y+15=x\left(3y+1\right)+15\left(y+1\right)=\left(x+15\right)\left(3y+1\right)\)

b.\(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3+x+y\right)\left(3-x-y\right)\)

c.\(x^3-5x^2+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x^2+1\right)\left(x-5\right)\)

d.\(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y+1\right)\left(x-y-1\right)\)

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

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

\(\left(5x-1\right)\left(1-5x+1\right)=0\)

\(\left(5x-1\right)\left(2-5x\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{2}{5}\end{array}\right.\)

\(9-\left(x-2\right)^2=0\)

\(\left(3+x-2\right)\left(3-x+2\right)=0\)

\(\left(1+x\right)\left(5-x\right)=0\)

\(\left[\begin{array}{nghiempt}x=-1\\x=5\end{array}\right.\)

\(3xy+x+15y-44=0\)

\(3y\left(x+5\right)+\left(x+5\right)-49=0\)

\(\left(x+5\right)\left(3y+1\right)=49\)

Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)

Có \(\left(x+5\right)\left(3y+1\right)=49\)

\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)

b tự lập bảng nhé~

Đặt \(xy-12x+15y\)là (*)

Từ phương trình (1) ta có \(x^2-3xy+2y^2+x-y=0\Leftrightarrow\left(x-y\right)\left(x-2y\right)+\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x=2y-1\end{cases}}\)

Với \(x=y\)thay vào (2) ta có \(x^2-2x^2+x^2-5x+7x=0\Leftrightarrow x=0\Rightarrow x=y=0\)

Thay \(x=y=0\)vào (*) ta thấy 0.0-12.0+15.0=0(tm)

Với \(x=2y-1\Rightarrow\left(2y-1\right)^2-2\left(2y-1\right)y+y^2-5\left(2y-1\right)+7y=0\)

\(\Leftrightarrow4y^2-4y+1-4y^2+2y+y^2-10y+5+7y=0\)

\(\Leftrightarrow y^2-5y+6=0\Leftrightarrow\left(y-2\right)\left(y-3\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}}\)

Với \(x=3;y=2\)thay vào (*)  ta thấy \(3.2-12.3+15.0=0\left(tm\right)\)

Với \(x=5;y=3\)thay vào (*)  ta thấy \(5.3-12.5+15.3=0\left(tm\right)\)

Vậy .....

2314654564

Hơn 170 điểm đã  là CTV rồi 

a/ \(x^4+4x^2-5\)

\(=\left(x^4+4x^2+4\right)-9\)

\(=\left(x^2+2\right)^2-9\)
\(=\left(x^2+2-3\right)\left(x^2+2+3\right)\)

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

b/ \(5x^3-5x^2y-10x^2+10xy\)

\(=\left(5x^3-10x^2\right)-\left(5x^2y-10xy\right)\)

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

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

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

Ta có: \(2x+3xy+15y=-15\\ \Leftrightarrow3y\left(x+5\right)+2\left(x+5\right)=-25\\ \Leftrightarrow\left(3y+2\right)\left(x+5\right)=-25\)

Nhưng mà \(x;y\in Z\) hay N để mình còn tính

chác là thuộc Z

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