Làm tính nhân

(4x³+3xy²-2y³).(3x²-5xy-6y²)

=12x⁵+12y⁵-20x⁴y-36x²y³-8xy⁴

Phân tích đa thức thành nhân tử

10x³+5x²y-10x²y-10xy²+5y³

=10x³-5x²y-10xy²+5y³

=5(2x³-x²y-2xy²+y³-)

3x² - 7x - 10

= 3x² + 3x - 10x - 10

= (3x² + 3x) - (10x + 10)

= 3x(x + 1) - 10(x + 1)

= (3x - 10)(x + 1)

5x³ - 5x²y - 10x² + 10xy

= 5x(x² - xy - 2x + 2y)

= 5x[(x² - xy) - (2x - 2y)]

= 5x[x(x - y) - 2(x - y)]

= 5x(x - 2)(x - y)

chúc bn hok tốt

nhớ ủng hộ tik cho mik nhé! ko lần sau mik ko giúp đâu

có j ko hiểu thì bn cứ bình luận ở dưới nhé

Đây là 1 hay 2 phần z bn

Trình bày cái đề cho đàng hoàng giúp.

a/ \(x^4+4x^2-5\)

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

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

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

b/ \(5x^3-5x^2y-10x^2+10xy\)

\(=\left(5x^3-10x^2\right)-\left(5x^2y-10xy\right)\)

\(=5x^2\left(x-2\right)-5xy\left(x-2\right)\)

\(=\left(x-2\right)\left(5x^2-5xy\right)\)

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

\(B=\frac{5x^2-10xy}{2\left(2y-x\right)}=\frac{5x\left(x-2y\right)}{-2\left(x-2y\right)}=\frac{5x}{-2}=\frac{-5x}{2}\)

2x^2-5x-7

=2x^2+2x-7x-7

=2x.(x+!) -7.(x+1)

(2x-7).(x+1)

x^2-y^2-5x+5y

=(x-y).(x+y)-5.(x-y)

=(x-y-5).(x+y)

5x^3-5x^2.y-10x^2+10xy

=5x^2.(x-y)-10x.(x-y)

=(5x^2-10x).(x-y)

=5x.(x-2).(x-y)

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)