Номер 1 а)25-36p^2c^2 б)(3x+1)^2-(4x+3)^2 в)х^2-9 г)2а(5а-10)+(2а-8)(3а+2)...

Question

Номер 1
а)25-36p^2c^2
б)(3x+1)^2-(4x+3)^2
в)х^2-9
г)2а(5а-10)+(2а-8)(3а+2)
д)(3х-5)(4х-5)-2х(2,5+1,5)
е)100а^4b^2c^2-121
ж)(a+b+c)^2-(a-b-c)^2
з)х^2-y^2
и)81а^4-1
Номер 2
a)3a^2-6ab-3b^2
б)ах^2+4ах+4а
в)-5а^2-10ab-5b^2
г)-3х^2+12х-12
д)а^2b-4abc+4bc^2
e)2х^3-4х+2
ж)-а^2+10ab-25b^2
з)-12х^3-12х^2-3х
^2 это значит в квадрате

Answer 1

A)25 - 36p²c²=5² - 6²p²c²=(5-6pc)(5+6pc)
б)(3х+1)²-(4х+3)²= (3х + 1 -(4х+3))(3х+1+4х+3) =
=(3х+1-4х-3)(7х+4) = (-х-2)(7х+4) = -(х+2)(7х+4)
в)х² - 9 = х² - 3² = (х-3)(х+3)
г)2а(5а-10)+(2а-8)(3а+2) = 10а² -20а  + 6а²-24а+4а-16 =
= 16а² -40а -16 = 8(2а²-5а -2)
д)(3х-5)(4х-5) -2х(2,5+1,5) = 12х² -15х -20х +25  - 2х *4=
=12х² - 35х + 25 - 8х = 12х² -43х +25  (???)
е)100а⁴b²c²  - 121 = 10²(a²)²b²c² - 11² = (10a²bc -11)(10a²bc+11)
ж)(а+b+c)²-(a-b-c)² =(a+b+c  -(a-b-c))(a+b+c +a-b-c) =
= (a+b+c-a+b+c)*2a = 2a(2b+2c) = 2a*2(b+c) = 4a(b+c) 
з)81a⁴  - 1 = (3²a²)²  - 1 = (3²a² -1)(3²a² +1) = (3a-1)(3a+1)(9a²+1)

№2.
а)3a²-6аb-3b²=-3(a²+2ab+b²) = -3(a+b)²
б)ах²+4ах +4а =а(х²+4х +4)= а(х² +2*2х +2²)=а(х+2)²
в)-5а²-10аb-5b² = -5(a²+2ab+b²)= -5(a+b)²
г)-3х²+12х-12 =-3(х² - 4х + 4) = -3(х² -2*2х +2²) = - 3(х-2)²
д)а²b-4abc+4bc² = b(a²-4ac +4c²) = b(a²-2*a*2c +(2c)² ) = b(a-2c)²
е)2х³-4х + 2 = 2(х³ -2х +1) = 2(x³-x  -x+1) = 2(x(x²-1) -(x-1)) =
= 2 (x(x-1)(x+1) - 1(x-1) ) = 2 (x-1)(x(x+1) - 1)  =
= 2(x-1)(x²+x-1)
з)-12х³-12х²-3х = -3х(4х² +4х + 1) =-3x((2x)²+2*2x*1 +1²)= -3x(2x+1)²