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)

\(A=\dfrac{x^2+2x-y^2-2y}{x^2-y^2}\\ =\dfrac{\left(x^2-y^2\right)+\left(2x-2y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y\right)+2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{x+y+2}{x+y}\)

b)

thay x=5,y=6 vào biểu thức A ta có

\(\dfrac{5+6+2}{5+6}=\dfrac{13}{11}\)

vậy A=13/11 kkhi x=5,y=6

a: \(A=\dfrac{\left(x+y\right)\left(x-y\right)+2\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{x+y+2}{x+y}\)

b: Khi x=5 và y=6 thì \(A=\dfrac{5+6+2}{5+6}=\dfrac{13}{11}\)

939393:3=313131 nhoa bẹn

a) \(A=\left(x-y\right).\left(x^2+x+y\right)-x.\left(2x^2+2y^3\right)\)

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

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

b) Ta thay \(x=-1;y=-5\)

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

\(=-\left(-1\right)^3-\left(-5\right)^2-2.\left(-1\right).\left(-5\right)^3\)

\(=1-25-250\)

\(=-274\)

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

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

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

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

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

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

\(A=100\)

Viết lại : 

a) \(M=\left(x+y\right)^3+2\left(x+y\right)^2\)

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

a) M=(x+y)³+2x²+4xy+2y²

     M=7³+(2x+2y)²=4(x+y)²=7³+4.7²=343+196=539

b)N=(x-y)³-x²+2xy-y²

    N=-5³-(x²-2xy+y²)=-125-(x-y)²=-125-(-5)²=-150

uiyiitiyiy

M=(x-y)²-2x-2y

= x²-2xy+y²-2(x+y)

= x²+2xy+y²-2(x+y)-4xy

= (x+y)²-2(x+y)-4xy

= 5²-2.5-4.6

= 25-10-24

= -9