a: \(x^4+2x^3+x^2=x^2\left(x+1\right)^2\)

b: \(5x^2+5xy-x-y\)

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

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

A = 3x² + 5y³ - 5xy² - 3xy + 2017

= (3x² - 3xy) + (5y³ - 5xy²) + 2017

= 3x(x - y) + 5y²(y - x) + 2017

= 3x(x - y) - 5y²(x - y) + 2017

= (x - y)(3x - 5y²) + 2017 

= 0 + 2017 = 2017 (Vì x - y) = 0

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

\(=8x^2-10xy+4x-10y+10y^2\)

\(3B=3\cdot\left(-3x^2+2xy-5y+y^2\right)\)

\(=-9x^2+6xy-15y+3y^2\)

\(5C=5\cdot\left(-x^2+3xy+2x+2y^2\right)\)

\(-5x^2+15xy+2x+2y^2\)

\(2A+3B\)

\(8x^2-10xy+4x-10y+10y^2-9x^2+6xy-15y+3y^2\)

\(=-x^2-4xy+4x-25y+13y^2\)

\(\left(2A+3B\right)-5C\)

\(=-x^2-4xy+4x-25y+13y^2-\left(\text{​​}\text{​​}-5x^2+6xy+10x+10y^2\right)\)

\(=-x^2-4xy+4x-25y+13y^2+5x^2-6xy-10x-10y^2\)

\(=4x^2-10xy-6x-25y+3y^2\)

vậy  2A+3B-5C=\(4X^2-10XY-6X-25Y+3Y^2\)

Ti ck nha

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

b: \(=xy\left(\dfrac{1}{5}+\dfrac{1}{3}\right)+xy^2\left(\dfrac{4}{3}-\dfrac{2}{5}\right)=\dfrac{8}{15}xy+\dfrac{14}{15}xy^2\)

d: \(=\dfrac{-4}{9}\cdot\dfrac{3}{2}\cdot xy^2\cdot xy^3=-\dfrac{2}{3}x^2y^5\)

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

a,=x^2+2x

c,=x^2-1

f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)