解不等式4a³－9a²＋6a－1≥0要过程
问题描述：
解不等式4a³－9a²＋6a－1≥0要过程
答
4a³-4a²-5a²+5a+a-1≥0
4a²(a-1)-5a(a-1)+(a-1)≥0
(a-1)(4a²-5a+a)≥0
(a-1)²(4a-1)≥0
过程就这样 拆开就看的清晰了 结果自己会算吧
答
4a³－9a²＋6a－1
=(4a^3-4a^2)-(5a^2-5a)+(a-1)
=(a-1)(4a^2-5a+1)
=(a-1)(a-1)(4a-1)
=(a-1)^2 (4a-1)>=0
∴a>=1/4
答
4a³－9a²＋6a－1≥04a³-4a²-(5a²-6a+1)≥04a²(a-1)-(5a-1)(a-1)≥0(a-1)(4a²-5a+1)≥0(a-1)(4a-1)(a-1)≥0(a-1)²(4a-1)≥0因为(a-1)²≥0所以4a-1≥0,解得a≥1/4或a=1...