Буду признателен, если решите:. Created by andreyshadow1. algebra-ru

Буду признателен, если решите:...

July 2022 1 12 Report

Буду признателен, если решите:

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svansen 1.a (x^2-3y)^2 = x^4-6x^2y+9y^2
   b (2a^2+1/3b^3)^2=4a^4+4/3a^2b^3+1/9b^6
   v (x^2-2y)(x^2+2y) = x^4 -4y^2
   g (3x-y)(y+3x) = 9x^2-y^2

2. (3a^2+2b)^2-(3a^2-b)^2 = (3a^2+2b-(3a^2-b))(3a^2+2b + 3a^2-b) = b*(6a^2+b)
_____________
   0.25a^4-3a^2b^2+9b^4 = (a^2/2)^2 -2*1/2*a^2*3*b^2 + (3b^2)^2 = (a^2/2 - 3b^2)^2
_________________
   x^2-6x+5
   36-20 = 16
   x1 = (6-4)/2 = 1
   x2 = (6+4)/2 = 5
   x^2-6x+5 = (x-1)(x-5)
   ______________
   x^2+4xy - 5y^2 = x^2 + 2*x*2y +4y^2 -9y^2 = (x+2y)^2-9y^2 = (x+2y-3y)(x+2y+3y) = (x-y)(x+5y)

3. 4(4-y^2)(y^2+4)-(5-y^3)^2 + (y^4+4y^2+16)(y^2-4) = 4(16-y^4) - (5-y^3)^2 + (y^6-2^6) = 2^6-4y^4+5^2+10y^3-y^6+y^6-2^6 = -4y^4+10y^3+25

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andreyshadow1 Огромное спасибо!!

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