Ta có :  \(N=\left(x-y\right)^3-x^2+2xy-y^2\)

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

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

Thay  \(x-y=-5\)vào ta được :

\(N=\left(-5\right)^3-\left(-5\right)^2\)

\(N=-125-25\)

\(N=-150\)

Vậy  \(N=-150\)với  \(x-y=-5\)

a) \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)

\(=7^3+2\left(x^2+2xy+y^2\right)\)

\(=343+2\left(x+y\right)^2\)

\(=343+2.7^2\)

\(=343+98=441\)

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

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

\(=-125-\left(-5\right)^2\)

\(=-125-25=-150\)

\(a,M=\left(x+y\right)^3+2x^2+4xy+2y^2\)

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

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

\(=\left(x+y\right)^2\left(x+y+2\right)=7^2.9=49.9=441\)

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

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

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

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

\(=\left(-5\right)^2\left(-5-1\right)=15.-6=-150\)

a) Ta có: A = (x + y)³ + 2x² + 4xy + 2y²

    A = 7³ + 2(x² + 2xy + y²)

   A = 343 + 2(x + y)²

 A = 343 + 2. 7²

  A = 343 + 98 = 441

b) B = (x - y)³ - x² + 2xy - y²

=> B = (-5)³ - (x² - 2xy + y²)

=> B = -125 - (x - y)²

=> B = -125 - (-5)²

=> B = -125 - 25 = -150

b) N= (x-y)³-x²+2xy-y²=(x-y)³-(x²-2xy+y²)=(x-y)³-(x-y)²

thay x-y=5 vào N ta được: N= -5³-5²=-150

a) đề sai k vậy bạn?

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

Bài 2:

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

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

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

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

P = x³ + y³ - x² - y² + 3xy( x + y ) - 2xy + 3( x + y ) + 10

= ( x³ + y³ ) - ( x² + 2xy + y² ) + 3xy( x + y ) + 3.5 + 10

= ( x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² ) - ( x + y )² + 3xy( x + y ) + 15 + 10

= [ ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² ) ] - 5² + 3xy( x + y ) + 25

= ( x + y )³ - 3xy( x + y ) - 25 + 3xy( x + y ) + 25

= 5³ = 125

Ta có : \(x^2+2x+y^2-2y-2xy+65\)

\(=\left(x-y\right)^2+2\left(x-y\right)+65\)

Mà \(x=y+5\)

\(\Rightarrow x-y=5\)

- Thay x  - y = 5 vào đa thức trên ta được :

\(=\left(x-y\right)^2+2\left(x-y\right)+65=100\)

Vậy ...