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) 2^x+1.3^y=12³

<=>2^x+1.3^y=(2²)³.3³

<=> 2^x+1=2⁶  và 3^y=3³

<=>x+1=6 và y=3

<=>x=5 và y=3

b) 10^x : 5^y=20^y

<=>10^x=20^y.5^y=100^y=(10²)^y

<=>x=2y (Nhiều số lắm chèn)

c) 2^x=4^y-1 

<=>2^x=2^y-2

<=>x=y-2

Mặt khác: 27^y=3^x+8

<=> 3^3y=3^x+8

<=>3y=x+8

<=>3y=(y-2)+8

<=>2y=6

<=>y=3

=>x=y-2=3-2=1

Sửa câu a xíu he

a) 2^x+1 . 3^y=12^x

<=>2^x+1.3^y=2^2x.3^x

<=>2^x+1=2^2x và 3^y=3^x

<=>x=y

và x+1=2x

<=>x=1 (và y=1)

=>Cặp (x;y)=(1;1)

a) Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=\left(2^2\right)^3\cdot3^3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2\cdot3\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)

Vậy: (x,y)=(5;3)

b) Ta có: \(10^x:5^y=20^y\)

\(\Leftrightarrow10^x=20^y\cdot5^y\)

\(\Leftrightarrow10^x=100^y\)

\(\Leftrightarrow x=2y\)