Question

已知x≤1,比较3x^3与3x^2-x+1的大小

Answer 1

解：∵[3x^3]-[3x^2-x+1]=3x^2(x-1)+(x-1)
=(x-1)(3x^2+1)
又 x≤1
从而 (x-1)≤0 (3x^2+1)>0
∴3x^3-3x^2+x-1≤0
则 [3x^3]-[3x^2-x+1]≤0
∴3x^3≤3x^2-x+1.

Answer 2

比较3x^3与3x^2-x+1的大小减法3x^3-（3x^2-x+1）=3x^3-3x^2+x-1=3x^2(x-1)+(x-1)=(x-1)(3x^2+1) x≤1,x-1≤0,3x^2+1>=1(x-1)(3x^2+1) ≤0, 3x^3≤3x^2-x+1

Answer 3

作差，得x^3-x^2+x-1=f（x）
则f'(x)=3x^2-2x+1
f'(x)=0则
x无解。即f'(x)＞0对于x∈R恒成立
所以f(x)在R上单调递增.
因为f(1)=0
X≥1
所以f(x)恒≥≥0
即X^3≥X^2-X+1,当x≥1时

Answer 4

3x^3=3x^2*x 则3x^3-3x^2=（X-1）3x^2 因为X小于1 所以（X-1）3x^2小于0 而-x+1大于0 所以3x^2-x+1大

Answer 5

3x³-（3x²-x+1）=(3x²+1)(x-1)
∵x≤1
∴(3x²+1)＞0
(x-1)≤0
∴ (3x²+1)(x-1)≤0
∴3x³≤3x²-x+1
当且仅当x=1时 相等