A=x^3 + y^3 + 3xy(x+y)
  =x+3x^y+3xy^2+y^3
  =(x+y)^3=2^3=8
B=x^2+2xy+y^2+4
  =(x+y)^2+4=4+4=8

C=x^3+y^3+3xy(x+y)+7(x+y)

  =(x+y)^3+7(x+y)
  =2^3+7.2
  =8+14=22

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

b, tương tự a

c, Sửa đề Cho a+b=1. Tính giá trị của các biểu thứ :A= a³+b³+3ab(a²+b²)+ 6a²b²(a+b)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay a+b=1 vào A ta có:

\(A=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)

d. \(B=x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1=\left(x+y\right)\left(x+y-4\right)+1\)

Thay x+y=3 vào B ta có:

\(B=3\left(3-4\right)+1=3.\left(-1\right)+1=-3+1=-2\)

a) Ta có: A = x³ + y³ + 3xy = (x + y)(x² - xy + y²) + 3xy = 1. (x² - xy + y²) + 3xy = x² - xy + y² + 3xy = x² + 2xy + y² = (x + y)² = 1² = 1

b)Ta có: B = x³ - y³ - 3xy = (x - y)(x² + xy + y²) - 3xy = 1. (x² + xy + y²) - 3xy = x² + xy + y² - 3xy = x² - 2xy + y² = (x - y)² = 1² = 1

d) Ta có : D = x³ + y³ + 3xy(x² + y²) + 6x²y²(x + y)

=> D = (x + y)(x² - xy + y²) + 3xy(x² + 2xy + y²) -  6x²y² + 6x²y²

=> D = x² - xy + y² + 3xy(x + y)² 

=> D = x² - xy + y² + 3xy.1²

=> D = x² + 2xy + y²

=> D = (x + y)² = 1² = 1

a) \(A=x^3+y^3+3xy\)

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

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

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

b) \(B=x^3-y^3-3xy\)

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

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

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

\(x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left[\left(x^2+2xy+y^2\right)-2xy\right]+6x^2y^2\)

\(=1-3xy+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)

\(=1-3xy+3xy\)

\(=1\)

1)a)x+y=60

<=>(x+y)^2=3600

<=>x^2+2xy+y^2=3600(1)

mà xy=35 nên 2xy=2.35=70

(1)<=>x^2+70+y^2=3600

<=>x^2+y^2=3530

<=>(x^2+y^2)^2=12460900

<=>x^4+2x^2.y^2+y^4=12460900(2)

mà xy=35 nên 2x.x.y.y=2450

(2)<=>x^4+y^4=123458450

 b)x+y=1

<=>(x+y)^3=1

<=>x^3+3x^2y+3xy^2+y^3=1

<=>x^3+y^3+3xy(x+y)=1

<=>x^3+y^3+3xy=1

=>M=1

x+y=1

<=>x^2+2xy+y^2=1(1)

B=x^3+y^3+3xy(x^2+y^2)+3xy(2xy)

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

=M.1=1(từ(1)

c)

x-y=1

<=>(x-y)^3=1

<=>x^3-3x^2y+3xy^2-y^3=1

<=>x^3-y^3-3xy(x-y)=1

<=>x^3-y^3-3xy=1

=>N=1

a. Có \(x+y=2\Rightarrow x^2+2xy+y^2=4\Rightarrow x^2+y^2=4-2.\left(-3\right)=10\)

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

\(=10^2-2.\left(-3\right)^2=82\)

b. Ta có \(x+y=1\Rightarrow x^2+y^2=1-2xy\)

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

\(=1.\left(1-2xy-xy\right)+3xy=1\)

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