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

Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

Đặt x²+x+1=t

Ta có: t(t+1)-12 = t²+t-12 = t²+4t-3t-12 = t(t+4) -3(t+4) =(t-3)(t+4) = (x²+x+1-3)(x²+x+1+4) =(x²+x-2)(x²+x+5).

= (x²+x+5)(x-1)(x+2)

\(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt:   \(x^2+x+1=t\)    Khi đó ta có:

\(A=t\left(t+1\right)-12\)

\(=t^2+t-12=\left(t-3\right)\left(t+4\right)\)

Thay trở lại đc:

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

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

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

(x^2+x+1)(x^2+x+2)-12

Đặt x^2+x+1= a ta có

=a^2+a-12

=a^2-3a+4a-12

=(a^2-3a)+(4a-12)

=a(a-3)+4(a-3)

=(a-3)(a+4)

thay x^2+x+1=a ta được

(x^2+x-2)(x^2+x+5)

                                         Bài làm :

Ta có :

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(x²+x+1)(x²+x+2)-12

=\(\left(x^2+x+\frac{3}{2}-\frac{1}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{1}{2}\right)-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{1}{4}-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{49}{4}\)
=\(\left(x^2+x+\frac{3}{2}-\frac{7}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{7}{2}\right)\)

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

\(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt:   \(x^2+x+1=t\)    Khi đó ta có:

\(A=t\left(t+1\right)-12\)

\(=t^2+t-12=\left(t-3\right)\left(t+4\right)\)

Thay trở lại đc:

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

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

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

Đặt : u=x²+x+1

Ta có : u(u+1)-12

=u²+u-12

=u²+4u-3u-12

=u(u+4)-3(u+4)

=(u+4)(u-3)

=(x²+x+1+4)(x²+x+1-3)

=(x²+x+5)(x²+x-2)

=(x²+x+5)[(x²+2x)-x-2]

=(x²+x+5)[(x²+2x)-(x+2)]

=(x²+x+5)(x+2)(x-1)

Nhận xét : x²+x+5>0 , với mọi x , nên không phân tích được nữa

Đặt x^2 + x+  1 = a => x^2 + x + 2 =a + 1 

Thay vòa ta có :

a( a+ 1 ) - 12 = a^2  + a - 12 = a^2 + 4a - 3a - 12 

=a (a+4) - 3 ( a+ 4 )

= ( a- 3 )(a+4)

Thây x^2 + x + 1 = a vào ta có 

(x^2 + x + 1 - 3 )(x^2 + x + 1 + 4 )

= ( x^2 + x - 2 )( x^ 2 + x + 5 )

đặt t=x²+x+1 ta được:

t.(t+1)-12

=t²-t-12

=t²+3x-4t-12

=t.(t+3)-4.(t+3)

=(t+3)(t-4)

thay t=x²+x+1 ta được:

(x²+x+4)(x²+x-3)

vậy (x² + x + 1) . (x² + x + 2) - 12=(x²+x+4)(x²+x-3)

Đặt x^2 + x +1 = a 

Thay vào ta có :

a(a+1) - 12 

= a^2 + a - 12 

= a^2 + 4a - 3a - 12 

= a(a+4 ) - 3 (a + 4 )

=(a- 3 )(a+4 )

Thay a = x^2 + x + 1 ta có :

= ( x^2 + x + 1 - 3 )(x^2 + x + 1 + 4 )

=(x^2 + x - 2 )(x^2 + x + 5 )