a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)

\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b² thành 27b³ vì mình nghĩ nhầm đề)

\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

a)  36-4a²+20ab-25b²

= 6^2 - (4a^2 - 20xb + 25b^2)

= 6^2 - (2a - 5b)^2

= [6 - (2a - 5b)] [6 + (2a - 5b)]

= (6 - 2a + 5b) (6 + 2a -5b)

a) \(6^2-\left(\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right)\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b) \(=\left(a+1\right)^3-\left(3b\right)^3\)

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

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

1/ a*b*(x^2+1)+x*(a^2+b^2)

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

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

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

3/ 9- x^2 - 2xy - y^2

= 9 - (x^2 + 2xy +y^2)

= 3^2 - (x+y)^2

= (3-x+y)(3+x+y)

4/ x^2 + y^2 + 2xy + yz + zx

= (x^2 + 2xy + y^2) + z(x+y)

= (x+y)^2 + z(x+y)

= (x+y+z)(x+y)

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

tớ lm câu a thui nha , tại khó quá ^^

a/ \(=3x^6+3x^5+6x^4+3x^3+3x^2-7x^5-7x^4-14x^3-7x^2-7x+3x^4+3x^3+6x^2+3x+1\)

\(=3x^2\left(x^4+x^3+2x^2+x+1\right)-7x\left(x^4+x^3+2x^2+x+1\right)+3\left(x^4+x^3+2x^2+x+1\right)\)

\(=\left(3x^2-7x+3\right)\left(x^4+x^3+x^2+x^2+x+1\right)\)

\(=\left(3x^2-7x+3\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)

\(=\left(3x^2-7x+3\right)\left(x^2+1\right)\left(x^2+x+1\right)\)

a) \(6x^3-12x^2y^2+6xy^3=6x.\left(x^2-2xy^2+y^3\right)\)

b) \(\left(x^2+4\right)^2-16=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\)

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

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

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

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

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

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

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

g) \(x^4-5x^2+4=x^4-4x^2+4-x^2=\left(x^4-4x^2+4\right)-x^2=\left(x^2-2\right)^2-x^2\)

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

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

k) \(\left(a^3-27\right)-\left(3-a\right)\left(6a+9\right)=\left(a-3\right).\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\)

\(\left(a-3\right)\left(a^2+3a+9+6a+9\right)=\left(a-3\right)\left(a^2+9a+18\right)\)

h) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

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

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

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

\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)

\(=\left(x-y\right)\left[\left(xy-zx\right)-\left(zy-z^2\right)\right]\)

\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)

\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Hai câu đầu tham khảo

Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath

c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)

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

\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)

Đặt \(x^2+5ax+4a^2=t\)

\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)

\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)

Mà \(x^2+5ax+4a^2=t\)

Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)

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

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

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

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