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问题: 高一数学题~~急~~

试比较a^3+8a与5a^2+4的大小

要有解题过程,越详细越好~~谢谢

解答:

(a^3+8a)-(5a^2+4)
=a^3+8a-5a^2-4
=a^3-a^2+a^2+8a-5a^2-4
=a^3-a^2-4a^2+8a-4
=a^2(a-1)-4(a^2-2a+1)
=a^2(a-1)-4(a-1)^2
=(a-1)[a^2-4(a-1)]
=(a-1)(a^2-4a+4)
=(a-1)(a-2)^2
当a<1时(a^3+8a)-(5a^2+4)=(a-1)(a-2)^2<0,a^3+8a<5a^2+4
当a=1时(a^3+8a)-(5a^2+4)=(a-1)(a-2)^2=0,a^3+8a=5a^2+4
当1<a<2时(a^3+8a)-(5a^2+4)=(a-1)(a-2)^2>0,a^3+8a>5a^2+4
当a=2时(a^3+8a)-(5a^2+4)=(a-1)(a-2)^2=0,a^3+8a=5a^2+4
当a>2时(a^3+8a)-(5a^2+4)=(a-1)(a-2)^2>0,a^3+8a5a^2+4
综上所述,
当a<1时,a^3+8a<5a^2+4
当a=1或a=2时,a^3+8a=5a^2+4
当1<a<2或a>2时,a^3+8a>5a^2+4