Sửa đề: \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)-28\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)

\(=\left(x^2+5x\right)^2-64\)

\(=\left(x^2+5x+8\right)\left(x^2+5x-8\right)\)

bạn học toán thầy tên Thắng à? Bài i chang bài mik đây ^v^ Tên bạn cx quen lắm 

(x-1)(x+2)(x+3)(x+6)-28

<=> (x-1)(x+6)(x+2)(x+3)-28

<=> (x²+5x-6)(x²​+5x+6)-28

đặt x²​+5x=t ta có:

<=>(t-6)(t+6)-28

<=>t²-36-28

<=>t²-64

<=>(t-8)(t+8)

thay t=x²+5x và đa thức ta có:

(x²​+5x-8)(x²​+5x+8)

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

Ta có: \(x^2-3x-28\)

\(=x^2-7x+4x-28\)

\(=x\left(x-7\right)+4\left(x-7\right)\)

\(=\left(x-7\right)\left(x+4\right)\)

7x² - 28 = 7(x² - 4) = 7(x - 2)(x + 2)

2x³ + 3x² - 18x - 27 = x²(2x + 3) - 9(2x + 3) = (2x + 3)(x² - 9) = (2x + 3)(x - 3)(x + 3)

\(\left(a+b\right)^3-3ab.\left(a+b\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]=\left(a+b\right)\left(a^2+b^2-ab\right)\)

`(a+b)^3-3ab(a+b)`

`=(a+b)(a+b)^2-3ab(a+b)`

`=(a+b)[(a+b)^2-3ab]`

`=(a+b)(a^2+2ab+b^2-3ab)`

`=(a+b)(a^2-ab+b^2)`

\(\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

 Thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

Biến đổi vế trái thành: 

a^3+b^3+c^3-3abc 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc 

<=>[(a+b)^3 +c^3] -3ab.(a+b+c) 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c) 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)

boc vai

Đặt \(a-b=x\) , \(b-c=y\) và \(c-a=z\)

\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)

Chắc bạn cùng biết  \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Vậy \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Chúc bạn học tốt.