3x除以2x-1大于等于1-搜答案-百度作业网

x-2除以6-x+2除以3等于1+x-1除以2两边乘6x-2-2(x+2)=6-3(x-1)-x-6=9-3x3x-x=9+62x=15x=7.5

9-2x>0,x

[-3,-1/3],作图,画出可行域即可.

先化简：去分母得3（a+x）大于等于2（2x+1）,然后去括号,然后就是3a+3x大于等于4x+2,移项的3a-2大于等于x,因为X大于等于1,所以3a-2大于等于1,所以3a大于等于3,所以a大于等

x/(2x-1)>=1x/(2x-1)-1>=0[x-(2x-1)]/(2x-1)>=0(-x+1)/(2x-1)>=0(x-1)/(2x-1)=0x-1

1/(x-1)+1/(x-4)-[1/(x-2)+1/(x-3)]=0通分(2x-5)/(x²-5x+4)-(2x-5)/(x²-5x+6)=0(2x-5)[1/(x²-

(|2X-1|-2)÷|X+3|≥0分母不等于0X≠-3|X+3|>0所以|2X-1|-2≥0|2X-1|≥22X-1≤-2,2X-1≥2X≤-1/2,X≥3/2综上X

x

若X>1,y>0,且满足xy=x^y,x/y=x^(3y)相乘得x2=x^(4y)y=0.50.5x=根号x0.25x2=xx=4x+y=4.5

x/2>x/33x>2x3x-2x>0x>0

(x-3)/2+3≥x(①)且1-3(x-1)＜8-x(②)解①式得：x-3+6≥2x3≥x即：x≤3解②式得：1-3x+3＜8-x4-8＜-x+3x2x＞-4x＞-2综合①②得：不等式组的解集是-2

∵x^2+x+1=(x+1/2)^2+3/4＞0则3x^2+2x+2≥M(x^2+x+1)即(3-M)x^2+(2-M)x+(2-M)≥0,∵对任意M恒成立,∴3-M＞0,且△=(2-M)^2-4(3

(3x-2)/(x-1)-2>=0(3x-1-2x+2)/(x-1)>=0(x+1)/(x-1)>=0即(x+1)(x-1)>=0分母不等于0所以x1

x+y>=0x-y>=0所以,两式相加,得2x>=0,x>=0又y>=-x,y

解题思路：去分母进行求解解题过程：答案见附件最终答案：略

(2+x)/2>=[(2x-1)/3]-23(2+x)>=2(2x-1)-2*63x-4x>=-6-2-12-x>=-20x

3>02x+1>02x>-1x>-1/2

y=(3x²+2x+2)/(x²+x+1)yx²+yx+y=3x²+2x+2(y-3)x²+(y-2)x+(y-2)=0x是实数,所以判别式大于等于0

x/(2x-1)≥1x/(2x-1)-1≥0(x-2x+1)/(2x-1)≥0(-x+1)/(2x-1)≥0(x-1)/(2x-1)≤01/2≤x≤1又∵2x-1≠0,即x≠1/2∴1/2

f（x）=x+2,x小于等于-1,增函数f（x）=2x,-1小于x小于2,增函数f（x）=x^2/2,x大于等于2增函数从图像上看,函数在此区间上既无最大值也无最小值