Respostas
respondido por:
1
3)
A = x⁶ - x⁵+ x - 1
B = x¹⁰ + 2x⁵ + 1
C = x¹⁰ - 1
fatorar
a) A = x⁶ - x⁵ + x - 1
A = x⁵(x - 1) +1(x - 1)
A = ( x⁵ + 1)(x - 1)
b) B = x¹⁰ + 2x⁵ + 1
B = x¹⁰ + x⁵ + x⁵ + 1
B = x⁵(x⁵ + 1)+ 1(x⁵ + 1)
B = (x⁵ + 1)(x⁵ + 1)
c) C = x¹⁰ - 1
C = x¹⁰ + X⁵ - X⁵ - 1
C = x⁵(x⁵ + 1) - 1(x⁵ + 1)
C = (x⁵ - 1)(x⁵ + 1)
d) M.M.C (A,C)=
A = ( x⁵ + 1)(x - 1)
C = (x⁵ - 1)(x⁵ + 1) ( x⁵ + 1)(x - 1),(x⁵ - 1)(x⁵ + 1) | (x - 1)
( x⁵ + 1) 1 (x⁵ - 1)(x⁵ + 1)| (x⁵- 1)
( x⁵ + 1) 1 1 (x⁵ + 1) | (x⁵ + 1)
1 1 1 1 /
então
M.M.C.(A,C) = (x -1)(x⁵ -1)(x⁵ +1)
e) M.M.C.(A,B,C) = ( x⁵ + 1)(x - 1)(x⁵ + 1)(x⁵ + 1) (x⁵ - 1)(x⁵ + 1)
M.M.C.(A,B,C) =( x⁵ + 1)(x - 1)(x⁵ - 1)
A = x⁶ - x⁵+ x - 1
B = x¹⁰ + 2x⁵ + 1
C = x¹⁰ - 1
fatorar
a) A = x⁶ - x⁵ + x - 1
A = x⁵(x - 1) +1(x - 1)
A = ( x⁵ + 1)(x - 1)
b) B = x¹⁰ + 2x⁵ + 1
B = x¹⁰ + x⁵ + x⁵ + 1
B = x⁵(x⁵ + 1)+ 1(x⁵ + 1)
B = (x⁵ + 1)(x⁵ + 1)
c) C = x¹⁰ - 1
C = x¹⁰ + X⁵ - X⁵ - 1
C = x⁵(x⁵ + 1) - 1(x⁵ + 1)
C = (x⁵ - 1)(x⁵ + 1)
d) M.M.C (A,C)=
A = ( x⁵ + 1)(x - 1)
C = (x⁵ - 1)(x⁵ + 1) ( x⁵ + 1)(x - 1),(x⁵ - 1)(x⁵ + 1) | (x - 1)
( x⁵ + 1) 1 (x⁵ - 1)(x⁵ + 1)| (x⁵- 1)
( x⁵ + 1) 1 1 (x⁵ + 1) | (x⁵ + 1)
1 1 1 1 /
então
M.M.C.(A,C) = (x -1)(x⁵ -1)(x⁵ +1)
e) M.M.C.(A,B,C) = ( x⁵ + 1)(x - 1)(x⁵ + 1)(x⁵ + 1) (x⁵ - 1)(x⁵ + 1)
M.M.C.(A,B,C) =( x⁵ + 1)(x - 1)(x⁵ - 1)
KatheCSomerhalder:
Muito obrigada!
respondido por:
0
MMC de letra '-' acho q divide por ela msm '--------------'
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