Задача 23692 Пожалуйста решите алгебру очень прошу я...
Условие
Заранее огромное спасибо 
Решение
ФОРМУЛА
(a+b)^2=a^2+2*a*b+b^2
1)
a)a=x; b=3
(x+3)^2=x^2+2*x*3+3^2=x^2+6x+9
б)a=8; b=y
(8+y)^2=8^2+2*8*y+y^2=64+16y+y^2
в) (p+a)^2= p^2+2*p*a+a^2=p^2+2pa+a^2
2)
a)(a-5)^2= a^2-2*a*5+5^2=a^2-10a+25
б)(7-c)^2= 7^2-2*7*c+c^2=49-14c+c^2
в) (x-10)^2=x^2-2*x*17+17^2=x^2-34x+289
3)
а)(3a-8)^2=(3a)^2-2*(3a)*8+8^2=9a^2-48a+64
б)(5x+6)^2=(5x)^2+2*(5x)*6+6^2=25x^2+60x+36
в)(2y-3)^2=(2y)^2-2*(2y)*3+3^2=4y^2-12y+9
4)
а)(3x+y)^2=(3x)^2+2*(3x)*y+y^2=9x^2+6xy+y^2
б)(2m-2n)^2=(2m)^2-2*(2m)*(2n)+(2n)^2=4m^2-8mn+4n^2
в)(-8x+a)^2=(-8x)^2-2*(8x)*a+a^2=64x^2-16ax+a^2
5)
а) (a^2-2)^2=(a^2)^2-2*a^2*2+2^2=a^4-4a^2+4
б)(b+c^2)^2=b^2+2*b*c^2+(c^2)^2=b^2+2bc^2+c^4
в)(x^2-y^2)^2=(x^2)^2-2*(x^2)*(y^2)+(y^2)^2=x^4-2x^2y^2+y^4
2.
(0,2x)^2+2*(0,2x)*2+2^2=0,04x^2+0,8x+4
(0,2x)^2-2*(0,2x)*2+2^2=0,04x^2-0,8x+4
(2y)^2+2*(2y)*(1/6)+(1/6)^2=4y^2+(2/3)y+(y^2/36)
(2y)^2-2*(2y)*(1/6)+(1/6)^2=4y^2-(2/3)y+(y^2/36)
(ab)^2+2*(ab)*4+4^2=a^2b^2+8ab+16
(ab)^2-2*(ab)*4+4^2=a^2b^2-8ab+16
(x^2)^2+2*(x^2)*(6a)+(6a)^2=x^4+12x^2a+36a^2
(x^2)^2-2*(x^2)*(6a)+(6a)^2=x^4-12x^2a+36a^2
(a^2b^2)^2+2*(a^2b^2)*8+8^2=a^4b^4+16a^2b^2+64
(a^2b^2)^2-2*(a^2b^2)*8+8^2=a^4b^4-16a^2b^2+64
3.
((a+b)+c)^2=(a+b)^2+2*(a+b)*c+c^2=
=a^2+2ab+b^2+2ac+2bc+c^2=
=a^2+b^2+c^2+2ab+2bc+2ac
((a-b)-c)^2=(a-b)^2-2*(a-b)*c+c^2=
=a^2-2ab+b^2-2ac+2bc+c^2=
=a^2+b^2+c^2-2ab+2bc-2ac
(x+y+z)^2=((x+y)+z)^2=(x+y)^2+2*(x+y)*z+z^2=
=x^2+2xy+y^2+2xz+2yz+z^2=
=x^2+y^2+z^2+2xy+2yz+2xz
(x-y-z)*(x-y-z)=(x-y-z)^2=((x-y)-z)^2=(x-y)^2-2*(x-y)*z+z^2=
=x^2-2xy+y^2-2xz+2yz+z^2=
=x^2+y^2+z^2-2xy+2yz-2xz
4.
(x-2y)^2=x^2-4xy+4y^2
3*(x-2y)^2=3*(x^2-4xy+4y^2)=3x^2-12xy+12y^2
9*(x-2y)^2=9*(x^2-4xy+4y^2)=[b]9x^2-36xy+36y^2[/b]
(3x-6y)^2=[b]9x^2-36xy+36y^2[/b]
(1/9)*(3x-6y)^2=(1/9)*(9x^2-36xy+36y^2)=x^2-4xy+4y^2
О т в е т.
(x-2y)^2=(1/9)*(3x-6y)^2
9*(x-2y)^2=(3x-6y)^2