Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với  - Hoc24

\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)

\(=\dfrac{\left(7x+2\right)\left(14x+4\right)}{3x^3y^3}\)

Quy đồng mẫu thức các phân thức sau :

a) 

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

b) \(x^2-6x-y^2+9=x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

c) \(7x^3y-14x^2y+7xy=7xy\left(x^2-2x+1\right)=7xy\left(x-1\right)^2\)

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

\(x^2-6x-y^2+9=\left(x^2-2.x.3+3^2\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(7x^3y-14x^2y+7xy=\left(7x^3y-7x^2y\right)-\left(7x^2y-7xy\right)=7x^2y.\left(x-1\right)-7xy.\left(x-1\right)\)

\(=\left(x-1\right).\left(7x^2y-7xy\right)=7xy.\left(x-1\right).\left(x-1\right)=7xy.\left(x-1\right)^2\)

d) \(16x^4y^2+2xy^5=2xy^2\left(8x^3+y^3\right)=2xy^2\left[\left(2x\right)^3+y^3\right]\)

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

e) \(5x^3-45x=5x\left(x^2-9\right)=5x\left(x^2-3^2\right)=5x\left(x-3\right)\left(x+3\right)\)