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

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

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

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

Ta có x + y = 3

=> (x + y)² = 9

<=> x²  + y² + 2xy = 9

<=> 2xy = 4

<=> xy = 2

Khi đó x³ + y³  = (x +  y)(x² - xy + y²) = 3.(5 - 2) = 9

b) Ta có x - y = 5 

<=> (x - y)² = 25

<=> x² - 2xy + y² = 25

<=> -2xy = 10

<=> xy = -5

Khi đó x³ - y³ = (x - y)(x² - xy + y²) = 5.(15 + 5) = 100 

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 

(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) 

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

a, x + y = 5

=> (x + y)^2 = 5^2

=> x^2 + 2xy + y^2 = 25

có xy = 4

=> x^2 + 2.4 + y^2 = 25

=> x^2 + y^2 = 17

+)vì x + y =5

=> (x+y)²=25

=> x²+2xy+y²=25

=>x²+y²+8=25 ( vì xy =4 )

=>x²+y²=17

x + y = 3

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

=> x^2 + 2xy + y^2 = 9

mà x^2 + y^2 = 5

=> 5 + 2xy = 9

=> 2xy = 4

=> xy = 2

x + y = 3

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

=> x^3 + 3x^2y + 3xy^2 + y^3 = 27

=> x^3 + y^3 + 3xy(x + y) = 27

xy = 2; x + y = 3

=> x^3 + y^3 + 3.2.3 = 27

=> x^3 + y^3 = 18 = 27

=> x^3 + y^3 = 9

Ta có: \(x+y=5\Rightarrow\left(x+y\right)^2=25\)

\(\Rightarrow x^2+2xy+y^2=25\)

\(\Rightarrow2xy+17=25\)\(\Rightarrow2xy=8\Rightarrow xy=4\)

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=17.\left(x+y\right)\left(x^2-xy+y^2\right)-4^2.5=17.5\left(17-4\right)-80=1025\)

`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.

`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`

`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`

`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.