Bài 1: 

a) (x+y)²=9²=81

=> x²+2xy+y²=81

=> x²+2.14+y²=81

=> x²+y²=53

=> x²-2xy+y²=81-2.14=25

=> (x-y)²=25

=> x-y=5 hoặc x-y=-5

b) Câu a đã tính được x²+y²=53

c) Ta có: x³+y³=(x+y)(x²-xy+y²)=9(53-14)=9.39=351

Bài 2: 

Ta có: \(x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1\)

Mà x+y=1

\(\Rightarrow1^2-4.1+1=-2\)

Bài 3: 

Ta có: (x+y)³=x³+3x²y+3xy²+y³ 

= x³+y³+3xy(x+y)

Mà x+y=1 => (x+y)³=x³+y³+3xy=1³=1

Bài 4: 

Ta có: \(\left(x+y\right)^2=4^2=16\)

\(\Rightarrow x^2+2xy+y^2=16\Rightarrow10+2xy=16\)

\(\Rightarrow2xy=6\Rightarrow xy=3\)

Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4.\left(10-3\right)\)

\(=4.7=28\)

Bài 5: 

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

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

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

Mấy bài này đầu hè làm hết rồi:))

Bài 1:

a) \(xy=14\Rightarrow x=\frac{14}{y}\)

Thay vào: \(\frac{14}{y}+y=9\)

\(\Leftrightarrow y^2+14-9y=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x=7\\y=2\end{cases}}\Rightarrow x-y=5\)

+ Nếu: \(\hept{\begin{cases}x=2\\y=7\end{cases}}\Rightarrow x-y=-5\)

b) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^2=81\)

\(\Leftrightarrow x^2+2xy+y^2=81\)

\(\Rightarrow x^2+y^2=81-2xy=81-2.14=53\)

c) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^3=9^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)

\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)

Bài 1:

a,A=x-y

  A²=(x-y)²

     =(x²+2xy+y²)-4xy

     =(x+y)²-4xy

     =9²-4.14 =25

b,x²+y²

=(x+y)²-2xy

=9²-2.12

=53

c,x³+y³

=(x+y)³-3xy(x+y)

=9³-3.14.9=351

Bài 8: Cho a+b= 1 nha ( mk thiếu đề)

Bài 1:

Theo bài ra ta có:

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

\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)

\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)

\(=25-10y+y^2+25-10x+x^2-4\)

\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)

\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)

\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)

\(=50-50+5^2-4-4\)

\(=25-8=17\)

Vậy giá trị của \(\left(x-y\right)^2\)là 17

ta có : M=2.(a^3  +b^3) -3.(a^2 + b^2)

       <=>M=2.(a+b)(a^2  -ab  +b^2)  - 3(a^2  +3b^2)

      <=>M=2(a^2  -ab  +b^2)  -3(a^2 +b^2)               vì a+b=1(gt)

      <=>M=-(a^2 +b^2 +2ab)

      <=>M=-(a+b)^2

      <=>M=-1  (vì a+b=1)

\(1,\left(x-1\right)\left(y-2\right)=2\left(x,y\in N\right)\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=2\\y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=1\\y-2=2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left(x;y\right)=\left\{\left(3;3\right);\left(2;4\right)\right\}\)

2/

2(x⁶+y⁶)-3(x⁴+y⁴)

=2[(x²)³+(y²)³ ] - 3x⁴-3y⁴

=2(x²+y²)(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2.1(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2x⁴-2x²y²+2y⁴-3x⁴-3y⁴

=-x⁴-2x²y²-y⁴

=-(x⁴+2x²y²+y⁴)

=-(x²+y²)

=-1

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