Question

已知a²+4a+1=0且2a³＋ma²＋2a/a的四次方＋ma²＋1＝3，求m的值

Answer 1

a^2 = -4a-1

(2a(-4a-1) + m(-4a-1) + 2a) / ((-4a-1)^2 + m(-4a-1) + 1)

= (-8(-4a-1)-2a + (-4am-m) + 2a) / (16a^2+8a+1 + (-4am-m) + 1)

= (32a+8 + (-4am-m) ) / (16(-4a-1)+8a+1 + (-4am-m) + 1)

= (32a+8 + (-4am-m) ) / (-64a-16+8a+1 + (-4am-m) + 1)

= (32a+8 + (-4am-m) ) / (-56a-14 + (-4am-m) ) = 3

32a+8 + (-4am-m) = - 168a - 42 - 12am - 3m

200a + 50 + 8am + 2m = 0

(8a+2)m = -200a-50

m = (-100a-25)/(4a+1)

= -25.