B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

\(A=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x^3+y^3\right)-2\left(x^2+y^2\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(\left(x+y\right)^2-2xy\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(x+y\right)^2+4xy+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(A=\left(5\right)^3-3xy\left(5\right)-2\left(5\right)^2+4xy+3xy\left(5\right)-4xy+3\left(5\right)+10\)

\(A=125-15xy-50+4xy+15xy-4xy+15+10\)

\(A=100\)

a) Thế x và y ta có:

\(-2.\left(-3\right)-5+11+3.\left(-3\right)\)

\(=6-5+11-9=3\)

b) Thế x và y ta có:

\(2.5-3.\left(-3\right)+5\left(5-\left(-3\right)\right)+15\)

\(=10+9+5\left(5+3\right)+15\)

\(=10+9+40+15=74\)

c) Thế x và y ta có:

\(4.\left(-3\right)-4\left(-3-2.5\right)-7\left(5-2\right)\)

\(=-12-4.\left(-13\right)-7.3\)

\(=-12+52-21=19\)

a: \(M=(x−y)(x2−xy)−x(x2+2y2)M=(x−y)(x2−xy)−x(x2+2y2)\)

\(=x3−x2y−x2y+xy2−x3−2xy2=x3−x2y−x2y+xy2−x3−2xy2\)

\(=−2x2y−xy2=−2x2y−xy2\)

Khi x=2 và y=-3 thì \(M=−2⋅22⋅(−3)−2⋅(−3)2M=−2⋅22⋅(−3)−2⋅(−3)2\)

\(=−8⋅(−3)−2⋅9=24−18=6=−8⋅(−3)−2⋅9=24−18=6\)

b: \(N=x(x2+xy+y2)−y(x2+xy+y2)N=x(x2+xy+y2)−y(x2+xy+y2)\)

\(=(x2+xy+y2)(x−y)=x3−y3=(x2+xy+y2)(x−y)=x3−y3\)

Khi x=4 và y=-1 thì \(N=43−(−1)3=64+1=65N=43−(−1)3=64+1=65\)

c: \(P=a2(a+b)−b(a2−b2)+2013P=a2(a+b)−b(a2−b2)+2013\)

\(=a2(a+b)−b(a−b)(a+b)+2013=a2(a+b)−b(a−b)(a+b)+2013\)

\(=(a+b)(a2−ba+b2)+2013=a3+b3+2013=(a+b)(a2−ba+b2)+2013=a3+b3+2013\)

Khi a=1 và b=-1 thì \(P=13+(−1)3+2013=2013P=13+(−1)3+2013=2013\)