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

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

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

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

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)

x^2+y^2=(x+y)^2-2xy

=5^2-2*3

=25-6

=19

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

=5^3-3*3*5

=125-9*5

=80

(x-y)^2=(x+y)^2-4xy=5^2-4*3=13

=>\(x-y=\sqrt{13}\)

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

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

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

Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

(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

\(1.\)

\(a)\)

\(x^2+y^2\)

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

\(=a^2-2b\)

\(b)\)

\(x^3+y^3\)

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

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

\(=a\left(a^2-3b\right)\)

\(=a^3-3ab\)

\(c)\)

\(x^4+y^4\)

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

\(=\left(a^2-2b\right)^2-2b^2\)

\(=a^4-4a^2b+2b^2\)

\(d)\)

\(x^5+y^5\)

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

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

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

\(=a^5-3a^3b-2a^3b+6ab^2-ab^2\)

\(=a^5-5a^3b+5ab^2\)