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

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

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

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

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

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

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

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

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

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

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

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 = 5x(x - y) - y(5x - y)

A = 5x² - 5xy - 5xy + y²

A = 5x² - 10xy + y² (1)

Thay x = -1; y = 3 vào (1), ta có:

5.(-1)² - 10.(-1).3 + 3² = 44

B = 4y(x² - 3xy + 3y²) - 2xy(2x - 6y - 3)

B = 4x²y - 12x² + 12y³ - 4x²y + 12xy² + 6xy

B = 12y³ + 6xy (1)

Thay x = 5; y = -1 vào (1), ta có:

12.(-1)³ + 6.5.(-1) = -42

C = 5x²(x - y²) + 3x(xy² - y) - 5x³ 

C = 5x³ - 5x²y² + 3x²y² - 3xy - 5x³ 

C = -2x²y² - 3xy (1)

Thay x = -2; y = -5 vào (1), ta có:

-2.(-2)².(-5)² - 3.(-2).(-5) = -230

D = 6x²(y² - xy + 2x²y) - 3xy(2xy - x² + 4x³)

D = 6x²y² - 6x³y + 12x⁴y - 6x²y² + 3x³y - 12x⁴y

D = -3x³y (1)

Thay x = 11; y = -1 vào (1), ta có:

-3.11³.(-1) = 3993

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

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

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

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~

a: Ta có: \(x^5-x^3+x^2-1\)

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

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

b: Ta có: \(5x^3-45x\)

\(=5x\left(x^2-9\right)\)

\(=5x\left(x-3\right)\left(x+3\right)\)

c: Ta có: \(16x^4y^2+2xy^5\)

\(=2xy^2\left(8x^3+y^3\right)\)

\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

d: Ta có: \(a^3-8+6a^2-12a\)

\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)

\(=\left(a-2\right)\left(a^2+8a+4\right)\)

e: Ta có: \(x^4+x^3+x+1\)

\(=x^3\left(x+1\right)+\left(x+1\right)\)

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

Đề bài là gì sao không ghi rõ?? 

a) \(3x^2-3xy-5x+5y\)

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

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

\(=2\left[\left(x+1\right)^2-y^2\right]\)

\(=2\left(x-y+1\right)\left(x+y+1\right)\)

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

d. ( x² - 2xy + y² ) ( x-y )
= ( x- y )² ( x- y )
= ( x - y )³

b, (x^2+xy+y^2) (x-y)
= ( x+ y )² ( x- y )
= ( x² - y ² ) ( x +y )