\(=\left(3x+15\right)^2-\left(x-7\right)^2=\left(4x+8\right)\left(2x+22\right)=8\left(x+2\right)\left(x+11\right)\)

(x-7)^2 = x^2-14x+49 

<=> 9x^2+90x+225 -x^2+14x-49 
= 8x^2+104x+176 
= 8(x^2+13+22) 
<=> 8(x+2)(x+11)

9x²+90x+225-(x-7)²

=(3x+15)²-(x-7)²

=(3x+15-x+7)(3x+15+x-7)

=(2x+22)(4x+8)

=2 (x+11)4 (x+2)

=8 (x+11)(x+2)

a) =( x²+y²-5)²-[2(xy-2)]²

⁼( x²+y²-5)²- (2xy-4)²

=(x²+y²-5+2xy-4)(x²+y²-5-2xy+4)

=[(x+y)²-9][(x-y)²-1]

phân tích tiếp HĐT 2 ở 2 thừa số

9x²+90x+225-(x-7)²

=(3x)²+2.3x.15+15²-(x-7)²

=(3x+15)²-(x-7)²

=[(3x+15)-(x-7)][(3x+15)+(x-7)]

=(3x+15-x+7)(3x+15+x-7)

=(2x+22)(4x+8)

=2(x+11).4(x+2)

=8(x+11)(x+2)

\(\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)

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

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

(x2+2x)+9x2+18x+20

=(x2+2x)+9(x2+2x)+20

Đặt t=x2+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x2+2x vào đc:

10(x2+2x+2)

a ,

b, 

c , 

bạn vô đường links sau để phân tích các bài toán đa thức thành nhân tử nha :

Cốc Cốc - Toán học | Trang chủ

\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right).\left(x^4-x^2y^2+y^4\right)\\ ---\\ 0,04-9x^2=\left(0,2\right)^2-\left(3x\right)^2=\left(0,2-3x\right)\left(0,2+3x\right)\\ ---\\ 32x^2-2\left(y-1\right)^2=2\left[16x^2-\left(y-1\right)^2\right]=2\left[\left(4x\right)^2-\left(y-1\right)^2\right]\\ =2\left(4x-y+1\right)\left(4x+y-1\right)\)

Câu 2 nha

\(a,x^4+2x^3+x^2\)

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

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

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

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

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

\(d,5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left[\left(x-y+2z\right)\left(x-y-2z\right)\right]\)

chiều lm tiếp cho 

\(A=4x^2+6x=2x\left(2x+3\right)\)

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)