2006-11-30 03:22:17 · 4 respostas · perguntado por Fernanda d 2 em Ciências e Matemática ➔ Matemática
f(x)=2x²-x+8 e f(3+p)-f(2-p)?
f(3+p) = 2(3+p)²-(3+p)+8
f(3+p) = 18+2p²+12(3+p)²-(3+p)+8
f(2-p) = 2(2-p)²-(2-p)+8
f(3+p)-f(2-p) =
2(3+p)²-(3+p)+8 - [2(2-p)²-(2-p)+8] =
2(3+p)²-(3+p)+8 - 2(2-p)²+(2-p)-8 =
2(3+p)²-3-p - 2(2-p)²+2-p =
2(3+p)²-1-2p - 2(2-p)² =
2(9+p²+6p) -1 -2p -2(4-4p+p²) =
18+2p²+12p -1 -2p-8+8p-2p² =
17+18p-8 =
18p + 9
Resposta: f(3+p)-f(2-p) = 18p + 9
₢
2006-11-30 03:49:21 · answer #1 · answered by Luiz S 7 · 0⤊ 0⤋
Na minha resolução...
o valor de f(3+p)-f(2-p) deu igual a 2p+9....
2006-11-30 18:09:03 · answer #2 · answered by lily-rose s2 2 · 0⤊ 0⤋
Fernanda,
Em primeiro lugar, calcularemos f(3+p).
f(3+p) = 2*((3+p)^2) -(3+p) + 8 = 2*(9 + 6*p + p^2) -(3+p) + 8
f(3+p) = 18 + 12*p + 2*(p^2) - 3 - p + 8
f(3+p) = 2*(p^2) + 11*p + 23
Em seguida, calcularemos f(2-p).
f(2-p) = 2*((2-p)^2) - (2-p) + 8 = 2*(4 - 4*p +p^2) - 2 + p + 8
f(2-p) = 8 - 8*p + 2*(p^2) - 2 + p + 8
f(2-p) = 2*(p^2) - 7*p + 14
Fazendo, por fim, f(3+p) - f(2-p), temos:
f(3+p) - f(2-p) = (2*(p^2) + 11*p + 23) - (2*(p^2) - 7*p + 14) =
= 2*(p^2) + 11*p + 23 - 2*(p^2) + 7*p - 14 =
= 11*p + 7*p + 23 - 14
f(3+p) - f(2-p) = 18*p + 9
2006-11-30 12:23:32 · answer #3 · answered by Verbena 6 · 0⤊ 0⤋
f(3+p)-f(2-p)=[2(3+p)^2-(3+p)+8]-[2(2-p)^2-(2-p)+8]=
2(9+6p+p^2)-3+p+8-2(4-4p+p^2)+2+p-8
18+12p+2p^2-3+p-8+8p-2p^2+2+p
9+22p
resposta: 9+22p
2006-11-30 11:29:39 · answer #4 · answered by Sara 3 · 0⤊ 0⤋