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大神们帮我看看第九题,谢谢了!
1个回答
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-1<=x<=3,x²-2x-3<=0
则-x²+2x+3-(2x-3)/(x-1)=0
-x³+2x²+3x+x²-2x-3-2x+3=0
-x³+3x²-x=0
x(x²-3x+1)=0
x是正数
则x=(3±√5)/2
都满足-1<=x<=3
x>3则是x²-2x-3-(2x-3)/(x-1)=0
x³-2x²-3x-x²+2x+3-2x+3=0
x³-3x²-3x+6=0
令y=x³-3x²-3x+6
令y'=3x²-6x-3
显然x>3时y'>0,递增
x=3,y=-3<0
而x=4时y>0
所以这里也有一个跟
一共3个,选C
则-x²+2x+3-(2x-3)/(x-1)=0
-x³+2x²+3x+x²-2x-3-2x+3=0
-x³+3x²-x=0
x(x²-3x+1)=0
x是正数
则x=(3±√5)/2
都满足-1<=x<=3
x>3则是x²-2x-3-(2x-3)/(x-1)=0
x³-2x²-3x-x²+2x+3-2x+3=0
x³-3x²-3x+6=0
令y=x³-3x²-3x+6
令y'=3x²-6x-3
显然x>3时y'>0,递增
x=3,y=-3<0
而x=4时y>0
所以这里也有一个跟
一共3个,选C
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