解不等式:(3x+2)/(x^2-3x+5)>2

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解不等式:(3x+2)/(x^2-3x+5)>2
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解不等式:(3x+2)/(x^2-3x+5)>2
解不等式:(3x+2)/(x^2-3x+5)>2

解不等式:(3x+2)/(x^2-3x+5)>2
移向通分
(-2x^2+9x-8)/(x^2-3x+5)>0
(2x^2-9x+8)/(x^2-3x+5)0
恒成立
只要2x^2-9x+8

(3x+2)/(x²-3x+5)>2
(3x+2)/(x²-3x+5)-2>0
(3x+2-2x²+6x-10)/(x²-3x+5)>0
(2x²-9x+8)/(x²-3x+5)<0
2(x²-9x/2+81/16+4-81/16)/(x²-3x+9/4+5-9/4)<0
[2(...

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(3x+2)/(x²-3x+5)>2
(3x+2)/(x²-3x+5)-2>0
(3x+2-2x²+6x-10)/(x²-3x+5)>0
(2x²-9x+8)/(x²-3x+5)<0
2(x²-9x/2+81/16+4-81/16)/(x²-3x+9/4+5-9/4)<0
[2(x-9/4)²-17/8]/[(x-3/2)²+11/4]<0
∵(x-3/2)²+11/4>0
∴2(x-9/4)²-17/8<0
(x-9/4)²<17/16
9/4-√17/4

收起

∵x²-3x+5=﹙x-3/2﹚²+11/4>0, ∴3x+2>2x²-6x+10,
2x²-9x+9<0, ﹙x-3﹚﹙2x-3﹚<0, ﹣3/2<x<3.

x^2-3x+5=(x-3/2)^2+11/4>0
故 3x+2>2x^2-6x+10 ==>0>2x^2-9x+8 ==>用求根公式就好了。。。