= ( x³ + 3x²y + 3xy² + y³ ) - 6xy - 3x² - 3y² + 3x + 3y + 2012

= ( x + y )³ - 3xy - 3x² - 3xy - y² + 3. ( x + y ) + 2012

= ( x + y )³ - 3x ( x + y ) - 3y .( x + y ) + 3.( x + y ) + 2012

= ( x + y )³ - 3.( x + y ) ( x + y ) + 3( x + y ) + 2012

= 101³ - 3.101² + 3.101 + 2012

= 1002013

Sửa đề: Các dấu bằng ở yêu cầu là dấu cộng.

1. Có: \(x+y=3\)

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

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

\(\Leftrightarrow x^2+y^2=9-2\cdot1=7\) (do \(xy=1\))

\(------\)

Lại có: \(x+y=3\)

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

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

\(\Leftrightarrow x^3+y^3+3\cdot1\cdot3=27\) (do x + y = 3; xy = 1)

\(\Leftrightarrow x^3+y^3=18\)

Ta có: \(x^2+y^2=7\)

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

\(\Leftrightarrow x^4+y^4+2\cdot\left(xy\right)^2=49\)

\(\Leftrightarrow x^4+y^4=49-2\cdot1=47\) (do xy = 1)

\(\text{a) Ta có:}xy=1\Rightarrow\hept{\begin{cases}2xy=2\\-2xy=-2\end{cases}}\)

\(\text{Ta lại có: }x^2+y^2=2\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=2+2=4\\x^2+y^2-2xy=2-2=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=4\\\left(x-y\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=\pm2\\x-y=0\end{cases}}}\)

\(\text{b) Ta có: }x+y=5\)

\(\Rightarrow\left(x+y\right)^2=25\)

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

\(\Rightarrow x^2+4+y^2=25\)

\(\Rightarrow x^2+y^2=21\)

\(\text{b) Ta có: }x^2+y^2=21\)

\(\Rightarrow x^2-2xy+y^2=21-2xy\)

\(\Rightarrow\left(x-y\right)^2=21-4\)

\(\Rightarrow\left(x-y\right)^2=17\)

\(\Rightarrow x-y=\pm\sqrt{17}\)

Ta có : \(A=x^2+y^2=x^2+2xy+y^2-2xy\)

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

Với \(x+y=3\) và \(xy=-10\)

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

\(A=9+20\)

\(A=29\)

Tương tự : \(B=x^3+y^3=\left(x+y\right)^3-3xy.\left(x+y\right)\)

\(B=\left(3\right)^3-3.\left(-10\right).3\)

\(B=117\)

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

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

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

\(=4^3+3\cdot5\cdot4+4^2\)

\(=64+16+60\)

=140

\(B=x^3-y^3+\left(x-y\right)^2=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)^2=\left(x-y\right)\left(x^2+xy+y^2+x-y\right)=\left(x-y\right)\left[\left(x-y\right)^2+\left(x-y\right)+3xy\right]=4\left(4^2+4+3.5\right)=140\)

\(P=\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}+2020=\dfrac{x^5+y^5}{\left(xy\right)^2}+2020=\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)-\left(xy\right)^2\left(x+y\right)}{\left(-2\right)^2}\)

\(=\dfrac{\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\left[\left(x+y\right)^2-2xy\right]-\left(-2\right)^2.5}{4}\)

\(=\dfrac{\left(-8+6.5\right)\left(25+4\right)-20}{4}=...\)

CÓ:     \(x^2+y^2=\left(x+y\right)^2-2xy=3^2-2.2=5\)

CÓ:     \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(5-2\right)=3.3=9\)

CÓ:     \(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=5^2-2.2^2=25-8=17\)

CÓ:     \(x^5+y^5=\left(x^4+y^4\right)\left(x+y\right)-x^4y-xy^4=3.17-xy\left(x^3+y^3\right)\)

\(=51-2.9=51-18=33\)

CÓ:     \(x^6+y^6=\left(x+y\right)\left(x^5+y^5\right)-xy^5-x^5y\)

\(=3.33-xy\left(x^4+y^4\right)=3.33-2.17\)

\(=99-34=65\)

\(x^2+y^2=\left(x+y\right)^2-2xy=3^2-2.2=9-4=5\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3.2.3=27-18=9\)

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

\(=3^4-4.2.5-3.2.2.2=81-40-24=17\)

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