a) 

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

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

\(C=x^5+y^5=\left(x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

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

\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-xy\left(x+y\right)\right)=a^5-5b\left(a^3-ab\right)\)

giup minh cau b o tren nha

(x+y)^2  =a^2

x^2 +2xy +y^2 =a^2

x^2+y^2 =a^2-2xy =a^2 -2b

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

             =a(a^2-2b-b)

            =a(a^2-3b)

            =a^3- 3ab

(x^2 +y^2)^2=(a^2-2b)^2  ( cái này tính cho x^4 + y^4)

tương tự như câu đầu tiên 

x^5+ y^5 (cái đó mình không biết)

sai con khi

\(1.\)

\(a)\)

\(x^2+y^2\)

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

\(=a^2-2b\)

\(b)\)

\(x^3+y^3\)

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

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

\(=a\left(a^2-3b\right)\)

\(=a^3-3ab\)

\(c)\)

\(x^4+y^4\)

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

\(=\left(a^2-2b\right)^2-2b^2\)

\(=a^4-4a^2b+2b^2\)

\(d)\)

\(x^5+y^5\)

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

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

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

\(=a^5-3a^3b-2a^3b+6ab^2-ab^2\)

\(=a^5-5a^3b+5ab^2\)

a, x + y = 3 => (x + y)² = 9 <=> x² + 2xy + y² = 9 <=>  5 + 2xy = 9 <=> 2xy = 4 <=> xy = 2

Ta có: x³ + y³ = (x + y)(x² - xy + y²) = 3 . (5 - 2) = 3 . 3 = 9

b, x - y = 5 => (x - y)² = 25 <=> x² - 2xy + y² = 25 <=> 15 - 2xy = 25 <=> -2xy = 10 <=> xy = -5

Ta có: x³ - y³ = (x - y)(x² + xy + y²) = 5 . (15 - 5) = 5 . 10 = 50 

1) Cho x+y=2 và x^2+y^2=10. Tính x^3+y^3. Giải

(x+y)^2=x^2+y^2+2xy => xy= -3 
x^3+y^3=(x+y)^3-3xy(x+y) = 26

2) Ta có: x^3+y^3 = (x+y)(x^2-xy+y^2) (1)

(x+y)^2=a^2

=> x^2 +2xy +y^2=a^2

=> b+2xy=a^2

=> xy=\(\frac{a^2-b}{2}\)

Thay (1) vào đó ta có:

x^3+y^3= (x+y)(x^2-xy+y^2) = a(b-\(\frac{a^2-b}{2}\)) = \(a\left(\frac{2b-a^2+b}{2}\right)=a.\frac{3b-a^2}{2}\)

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

Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=2^2-2xy=4-2xy=10\Rightarrow2xy=-6\Rightarrow xy=-3\)

Vậy: \(x^3+y^3=2\left(10+3\right)=2.13=26\)

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

a) \(\left(x+y\right)^2=x^2+y^2+2xy\Rightarrow4=10+2xy\Leftrightarrow xy=-3\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3+3.3.2=26\)

b) \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow m^2=n-2xy\Leftrightarrow xy=\frac{n-m^2}{2}\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=m^3+3.m.\frac{n-m^2}{2}=\frac{3mn}{2}-\frac{m^3}{2}\)

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

\(=3^3-3\cdot3\cdot\left(-2\right)\)

\(=27+18=45\)

b: \(\left(x^3+y^3\right)^2=45^2=2025\)

a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)

\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)

\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)

b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)​​

​\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)​​

\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)