Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử

\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)

\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)

\(=12x^2-6x=6x\left(2x-1\right)\)

Ta có:  9 x 2 - 1 = ( 3 x ) 2 - 1 2 = ( 3 x - 1 ) ( 3 x + 1 )

Ta có:  9 x 2 - 1 = ( 3 x ) 2 - 1 2 = ( 3 x - 1 ) ( 3 x + 1 )

`9x^2-4=(3x)^2-2^2=(3x-2)(3x+2)`

\(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)

ADHĐT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\(9x^2-4y^2\)

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

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

(x + y)2 – 9x2 = (x + y)2 – (3x)2

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

= (4x + y)(-2x + y)

giúp em với ạ TvT em đang cần gấp lắm huhu

b)

Sửa đề: \(125a^3+75a^2+15a+1\)

Ta có: \(125a^3+75a^2+15a+1\)

\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)

\(=\left(5a+1\right)^3\)

c) Ta có: \(64-96a+48a^2-8a^3\)

\(=-\left(8a^3-48a^2+96a-64\right)\)

\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)

\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)

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

\(=-8\left(a-2\right)^3\)

\(a,=\left(3x+1\right)^2-y^2=\left(3x-y+1\right)\left(3x+y+1\right)\\ b,=x\left(x^2-5x+6\right)=x\left(x^2-2x-3x+6\right)=x\left(x-2\right)\left(x-3\right)\)