Question

Se A = 4x²y^5 , B = 6x³y³a e C = 10x²y^4b, determine o m.d.c (máximo divisor comum):

(A) m.d.c. (A,B,C) = 24abx^7y^9
(B) m.d.c. (A,B,C) = 40abx^7y^9
(C) m.d.c. (A,B,C) = 24x^5y^8
(D) m.d.c. (A,B,C) = 2x³y^5
(E) m.d.c. (A,B,C) = 2x²y³

Answer 1

Oi Gabriel 

mdc(4, 6, 10) = mdc(2^2, 2*3, 2*5) = 2

mdc(x², x³, x²) = x²

mdc(y⁵, y³, y⁴) = y³ 

mdc(A, B, C) = 2x²y³ (letra E)

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