a: Ta có: \(A=10x^2+20xy+10y^2-90\)

\(=10\left(x^2+2xy+y^2-9\right)\)

\(=10\left(x+y-3\right)\left(x+y+3\right)\)

b: Ta có: \(B=x^3y-3x^2y-4xy+12y\)

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

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

c: Ta có: \(C=125x^3-10x^2+2x-1\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)

\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)

Chọn A

Thực hiện phép chia đa thức cho đa thức ta được:

(10x2 + 23x – 5) : (2x + 5) = 5x - 1

Bài 1:
\(\left(x^2-y\right)\left(3x+y^2\right)-\left(6x^4y-2xy^4\right):2xy\)

\(=3x\cdot x^2+y^2\cdot x^2-y\cdot3x-y\cdot y^2-6x^4y:2xy+2xy^4:2xy\)

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

\(=x^2y^2-3xy\) 

Bài 2:

a) \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)

\(=10x^2\left(2x-y\right)-6xy\left(2x-y\right)\)

\(=2x\left(2x-y\right)\left(5x-3y\right)\)

b) \(x^2-2x+1-y^2\)

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

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

c) \(x^2-8x+12\)

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

\(=\left(x-4\right)^2-2^2\)

\(=\left(x-6\right)\left(x+2\right)\)

\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)

\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)

A

D bạn nhé

a) \(24x^2-4xy\)

\(=4x\left(6x-y\right)\)

b) \(5x^3-10x^2+5x-20xy^2\)

\(=5x\left(x^2-10x+5-20y^2\right)\)

nhầm câu b nha 

\(5x^3-10x^2+5x-20xy^2\)

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

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

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

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

a: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{x^3-10x^2+25x}=\dfrac{\left(x-5\right)\left(x+5\right)}{x\left(x-5\right)^2}=\dfrac{x+5}{x\left(x-5\right)}\)

b: \(=\dfrac{y-2}{\left(y-2\right)\left(y+1\right)}=\dfrac{1}{y+1}\)

Đáp án C

Ta có:

Vậy hệ phương trình đã cho có 2 nghiệm.

\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)