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 

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

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

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

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

`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`.

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

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

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=13.19-\left(-6\right)^2.1=211\)

b)  \(x^2+y^2=\left(x-y\right)^2+2xy=1^1+2.6=13\)

    \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+3.6.1=19\)

   \(x^5-y^5=\left(x^2+y^2\right)\left(x^3-y^3\right)+x^2y^2\left(x-y\right)=13.19+6^2.1=283\)

a) Ta có x + y = 25

=> (x + y)² = 625

=> x² + y² + 2xy = 625

=> x² + y² + 10 = 625

=> x² +y² = 615

b) Ta có x + y = 3

=> (x + y)³ = 27

=> x³ + 3x²y + 3xy² + y³ = 27

=> x³ + y³ + 3xy(x + y) = 27

=> x³ + y³ + 9xy = 27 

Lại có x + y = 3

=> (x + y)² = 9

=> x² + y² + 2xy = 9

=> 2xy = 4

=> xy = 2

Khi đó x³ + y³ + 9xy + 27

=> x³ + y³ + 18 = 27

=> x³ + y³ = 9

c) Ta có x - y = 5

=> (x - y)² = 25

=> x² + y² - 2xy = 25

=> 2xy = -10

=> xy = -5

Khi đó : x³ - y³ = (x - y)(x² + xy + y²) = 5(15 - 5) = 5.10 = 50

Bài 4.

a) x² + y² = x² + 2xy + y² - 2xy

= ( x² + 2xy + y² ) - 2xy

= ( x + y )² - 2xy

= 25² - 2.136

= 625 - 272

= 353

b) x + y = 3

⇔ ( x + y )² = 9

⇔ x² + 2xy + y² = 9

⇔ 5 + 2xy = 9 ( gt x² + y² = 5 )

⇔ 2xy = 4

⇔ xy = 2

x³ + y³ = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

= ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

= ( x + y )³ - 3xy( x + y )

= 3³ - 3.2.3

= 27 - 18

= 9 

a) Ta có x + y = 25

=> (x + y)² = 625

=> x² + y² + 2xy = 625

=> x² + y² + 2.136 = 625

=> x² + y² + 272 = 625

=> x² + y² = 353

a)  \(x+y=3\)

\(\Rightarrow\)\(\left(x+y\right)^2=9\)

\(\Leftrightarrow\)\(x^2+y^2+2xy=9\)

\(\Leftrightarrow\)\(2xy=4\)  do x² + y² = 5

\(\Leftrightarrow\)\(xy=2\)

   \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3.2.3=9\)

b) bạn làm tương tự

\(a,x+y=3\Rightarrow\left(x+y\right)^2=9\Rightarrow x^2+2xy+y^2=9\Rightarrow2xy=4\Leftrightarrow xy=2\)

Vì \(\left(x+y\right)=3\Rightarrow\left(x+y\right)^3=27\)

\(\Rightarrow x^3+3x^2y+3xy^2+y^3=27\)

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\)

\(\Rightarrow x^3+y^3+3.2.3=27\)

\(\Rightarrow x^3+y^3=27-18=9\)

\(b,x-y=5\Rightarrow\left(x-y\right)^2=25\Rightarrow x^2-2xy+y^2=25\Rightarrow2xy=-10\Leftrightarrow xy=-5\)

\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5.10=50\)

Bài làm:

a) Ta có: \(x+y=3\Leftrightarrow\left(x+y\right)^2=3^2\Leftrightarrow x^2+2xy+y^2=9\)

\(\Leftrightarrow2xy+5=9\)

\(\Leftrightarrow2xy=4\)

\(\Rightarrow xy=2\)

Từ đó ta có: \(x^3+y^3\)

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

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

\(=3.3=9\)

b) Ta có: \(x-y=5\Leftrightarrow\left(x-y\right)^2=25\Leftrightarrow x^2-2xy+y^2=25\)

\(\Leftrightarrow15-2xy=25\)

\(\Leftrightarrow2xy=-10\)

\(\Rightarrow xy=-5\)

Từ đó ta có: \(x^3-y^3\)

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

\(=5\left(15+10\right)\)

\(=5.25=125\)

Bài 1.

A = x² + 2xy + y² = ( x + y )² = ( -1 )² = 1

B = x² + y² = ( x² + 2xy + y² ) - 2xy = ( x + y )² - 2xy = (-1)² - 2.(-12) = 1 + 24 = 25

C = x³ + 3xy( x + y ) + y³ = ( x³ + y³ ) + 3xy( x + y ) = ( x + y )( x² - xy + y² ) + 3xy( x + y )

                                                                                  = -1( 25 + 12 ) + 3.(-12).(-1)

                                                                                  = -37 + 36

                                                                                  = -1

D = x³ + y³ = ( x³ + 3x²y + 3xy² + y³ ) - 3x²y - 3xy² = ( x + y )³ - 3xy( x + y ) = (-1)³ - 3.(-12).(-1) = -1 - 36 = -37

Bài 2.

M = 3( x² + y² ) - 2( x³ + y³ )

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

= 3( x² + y² ) - 2( x² - xy + y² )

= 3x² + 3y² - 2x² + 2xy - 2y²

= x² + 2xy + y²

= ( x + y )² = 1² = 1

Bài 3.

x + y = -3

<=> ( x + y )² = 9

<=> x² + 2xy + y² = 9

<=> 29 + 2xy = 9

<=> 2xy = -20

<=> xy = -10

x³ + y³ = ( x³ + 3x²y + 3xy² + y³ ) - 3x²y - 3xy² = ( x + y )³ - 3xy( x + y ) = ( -3 )³ - 3.(-10).(-3) = -27 - 90 = -117

Bài 4.

x - y = 5

<=> ( x - y )² = 25

<=> x² - 2xy + y² = 25

<=> 15 - 2xy = 25

<=> 2xy = -10

<=> xy = -5

x³ - y³ = ( x³ - 3x²y + 3xy² - y³ ) + 3x²y - 3xy² = ( x - y )³ + 3xy( x - y ) = 5³ + 3.(-5).5 = 125 - 75 = 50