|3x+2|≥|2x+a|
<==>9x^2+12x+4≥4x^2+4ax+a^2
<==>5x^2+4(3-a)x+(4-a^2)≥0
要使x∈R恒成立,即使判别式△≤0.
也即[4(3-a)]^2-20(4-a^2)≤0.
<==>9a^2-24a+16≤0
<==>(3a-4)^2≤0
<==>a=4/3.
若不等式|3x+2 |≥|2x+a |对x∈R恒成立,则实数a的取值范围是?
若不等式|3x+2 |≥|2x+a |对x∈R恒成立,则实数a的取值范围是?
数学人气:737 ℃时间:2019-09-01 10:49:45
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