`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`.

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

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

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

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

(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 + 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

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

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

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

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

x^3+y^3=(x+y)^3-3xy(x+y)

=5^3-3*6*5

=125-90

=35

\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)

ĐKXĐ: \(x\ne y\)

a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)

b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)

\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)