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)

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

Bài 1: 

a,  \(x^2\) +2\(x\) = 0

     \(x.\left(x+2\right)\) = 0

     \(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

      \(x\) \(\in\) {-2; 0}

b, (-2.\(x\)).(-4\(x\)) + 28  = 100

      8\(x^2\)           + 28  = 100

        8\(x^2\)                   = 100 - 28

        8\(x^2\)                   = 72

          \(x^2\)                  = 72 : 8

          \(x^2\)                   = 9

           \(x^2\)                  = 3²

          |\(x\)|                  = 3

          \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\) 

Vậy \(\in\) {-3; 3}

c, 5.\(x\) (-\(x^2\)) + 1 = 6

   - 5.\(x^3\)       + 1 = 6

   5\(x^3\)                 = 1 - 6

   5\(x^3\)                 = - 5

    \(x^3\)                  =  -1

    \(x\)                    =  - 1

d, 3\(x^2\) + 12\(x\) = 0

   3\(x.\left(x+4\right)\) = 0

   \(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-4; 0}

e, 4.\(x.3\) = 4.\(x\)

    12\(x\) - 4\(x\) = 0

      8\(x\)          = 0

         \(x\)        = 0

a)

\(C=x^2+y^2=\left(x-y\right)^2+2xy=3^2+2.18=45.\)

b)

\(D=x^3+y^3=\left(x-y\right)^3+3x^2y-3xy^2=\left(x-y\right)^3+3xy\left(x-y\right)=3^3+3.18.3=189\)

Đề sai rồi, không thể tồn tại x; y sao cho \(\left\{{}\begin{matrix}x+y=3\\xy=5\end{matrix}\right.\) được

Vì \(\left(x+y\right)^2\ge4xy;\forall x;y\) nên \(3^2>4.5\) là vô lý

a: \(x^2+y^2=\left(x+y\right)^2-2xy=3^2-2\cdot5=-1\)

b: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3\cdot3\cdot5=-18\)

b) \(xy+3x-2y=11\)

\(xy+3x-2y-6=11-6\)

\(xy+3x-2y-6=5\)

\(\left(xy+3x\right)-\left(2y+6\right)=5\)

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

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

\(\Rightarrow5=\left(-1\right)\left(-5\right)=1\cdot5\)

Bạn tự lập bảng mà thử nghiệm nhé

Ngu vcl

a) \(xy+3x-7y=21\)

\(xy+3x-7y-21=21-21\)

\(xy+3x-7y-21=0\)

\(\left(xy+3x\right)-\left(7y+21\right)=0\)

\(x\left(y+3\right)-7\left(y+3\right)=0\)

\(\left(x-7\right)\left(y+3\right)=0\)

\(\orbr{\begin{cases}x-7=0\\y+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\y=-3\end{cases}}\)

Thử

a) \(\begin{cases}x+5=\left(-11,-1,1,11\right)\\y+3=\left(-1,-11,11,1\right)\end{cases}\Leftrightarrow\hept{\begin{cases}x=\left\{-16,-6,-4,6\right\}\\y=\left\{-4,-14,8,-2\right\}\end{cases}}\)

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

= ( 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