Y^3dy=x^2dx y=3 при x=2

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Математика

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8 июля, 03:48

Y^3dy=x^2dx y=3 при x=2

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Вестита · Answer 1

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

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

(y+10) (y-2) - 4y (2 - 3y) = y^2+8y-20-8y+12y^2=13y^2-20

(a-4) (a+9) - 5a (1-2a) = a^2+5a-36-5a+10a^2=11a^2-36

(2b-3) (3b+2) - 3b (2b+3) = 6b^2-9b+4b-6-6b^2-9b=-14b-6

(3a-1) (2a-3) - 2a (3a+5) = 6a^2-2a-6a+4-6a^2-10a=-18a+4

(m+3) ^2 - (m-2) (m+2) = m^2+6m+9-m^2+4=5m+13

(a-1) ^ - (a+1) (a-2) = a^2-2a+1-a^2-a-2=-3a-1

(c+2) (c-3) - (c-1) ^2=c^2-c-6-c^2+2c-1=c-7

(y-4) (y+4) - (y-3) ^=y^2-16-y^2+6y-9=6y-25

(a-2) (a+4) - (a+1) ^ = a^2+2a-8-a^2-2a-1=-9

(b-4) (b+2) - (b-1) ^=b^2-2b-8-b^2+2b-1=-9