\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

 =(y+x-6)(y+x-2)

Đặt \(x+y=u\)

Biểu thức trở thành \(u^2-8u+12\)

\(=u^2-2u-6u+12\)

\(=u\left(u-2\right)-6\left(u-2\right)\)

\(=\left(u-6\right)\left(u-2\right)\)

Thay ngược trở lại, ta được:

\(\left(x+y\right)^2-8\left(x+y\right)+12=\left(x+y-6\right)\left(x+y-2\right)\)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

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

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

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

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

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

a) (x - 1)(x + l)(x - 2)(x - 4).      b) (x - 2)( x 2  + 4).

c) 2y(3 x 2   +   y 2 ).                          d) 2(x + y + z) ( a   -   b ) 2 .

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

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

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

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

\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)

b. \(x^3-2x^2+4x-8\)

\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)

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

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

c. \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3\)

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

d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)

\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)

\(=2\left(a-b\right)^2\left(x+y+z\right)\)

a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)

b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)

\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)

a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)

b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)

=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)

a) Biến đổi x 3   =   x 2 .x, phân tích thành x( x 2  + 2).

b) Tương tự a) phân tích thành 3(x – 2y).

c) Nhân tử chung 5(x + 3y) phân tích thành 5(x + 3y)(1 – 3x).

d) Thực hiện biến đổi y – x = -(x – y), xuất hiện nhân tử chung là (x – y), phân tích thành (x – y)(3 + 5x).