解不等式4x 1分之3x-1大于等于1
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是不是这个5/(2x+3)-1≥2(x-1)5/(2x+3)≥2x-1∵5/(2x+3)-1≥2(x-1)存在∴2x+3≠0∴5≥4x^2+4x-34x^2+4x-8≤0x^2+x-2≤0-2≤x≤1
-2x+1>-11-2x>-12∴x=2xx>=1∴1
3x-6≥x-42x≥2x≥1两边乘32x+1>3x-3x
写纸上给你参考下吧.再答:
(3-x)/(x+1)-2>=0(3-x-2x-2)/(x+1)>=0(-3x+1)/(x+1)>=0(3x-1)/(x+1)再问:(3x-1)/(x+1)=03x-1
分情况讨论,a大于零,a<零,a等于零
两端同时乘以2:x-3+6=2x+2合并同类项:x
(3-x)/(x+1)-2>=0(3-x-2x-2)/(x+1)>=0(-3x+1)/(x+1)>=0(3x-1)/(x+1)
(2-x)/4>3-x/22-x>12-6x5x>10x>2
解x-3(x-2)>4x-3x+6>4-2x>-2x
1>xx≥-2故-2,-1,0
{8-x>3x,(5x+1)/3≥x-1{x<2,5x+1≥3x-3{x<2,2x≥-4{x<2,x≥-2-2≤x<2
x/2+(x+1)/3>0(1)x+(5a+4)/3>(4(x+1)+a)/3(2)由(1)式得x>-2/5,由(2)式得3x+5a+4>4x+4+a,x
x(x-1)(x-2)³(x+1)(x-1)(x-1)(x²+x+1)/[(3-4x)(x²+x+2)]>0x(x-1)³(x-2)³(x+1)(x&
1.由x/2>-1得,x>-2由2x+1≥5(x-1)得,x≤2∴-25(x-1)得,x
第一个是对的:x-1>0x>1,所以有无数个解.第二个是错的:2x-3≤02x≤3x≤二分之三而不是≥三分之二.这是八下第一章内容...我也是放假后预习才懂的O(∩_∩)O,
可得、x≠4当x<4时:x+2≥4-xx≥11≤x<4当x>4时:x+2≤4-xx≤1不成立所以1≤x<4
两边同乘以十再算,结果是x大于十三分之十五
(4x-1)/2≥(1-5x)/6,去分母,不等式两边同时乘6,得:3(4x-1)≥1-5x,去括号:12x-3≥1-5x,移项:12x+5x≥1+3,合并同类项:17x≥4x≥4/17
两边乘以6得3(3x-4)-2(x-1)>=129x-12-2x+2>=127x>=22x>=22/7