Question

已知a²-a-1=0且2a的四次方-3xa²+2/a³+2xa²-a=-93/112求x的值，最好今天。要有解题的

属于初二奥数

Answer 1

a^2-a-1=0
a^2-1=a
a^2=a+1
(2a^4-3xa^2+2)/(a^3+2xa^2-a)=-93/112
(2a^4-3xa^2+2)/(a^3+2xa^2-a)
=[a^2(2a^2-3x)+2] / [a(a^2-1+2xa)]
={(a+1)[2(a+1)-3x]+2} / [a(a+2xa)]
={2(a+1)^2-3(a+1)x]+2} / [a^2(1+2x)]
={2a^2+4a+2-3(a+1)x]+2} / [(a+1)(1+2x)]
={2(a+1)+4a+2-3(a+1)x]+2} / [(a+1)(1+2x)]
={2(a+1)+4(a+1)-3(a+1)x]} / [(a+1)(1+2x)]
={(a+1)(6-3x)} / [(a+1)(1+2x)]
=3(2-x) / (1+2x)=-93/112
(2-x) / (1+2x)=-31/112
112(2-x)=-31(1+2x)
224-112x=-31-62x
50x=255
x=255/50=51/10