log2(x-1/x)≥1 怎么算.急- 答案是x>1,x
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/02 07:13:15
log2(x-1/x)≥1 怎么算.急- 答案是x>1,x
log2(x-1/x)≥1 怎么算.急
- 答案是x>1,x
log2(x-1/x)≥1 怎么算.急- 答案是x>1,x
log2(x-1/x)≥log2(2)
(x-1)/x≥2
(x+1)/x≤0
则:{x|-1≤x
log2(x-1/x)≥log2(2)
x-1/x≥2
x-1/x-2≥0
(x²-2x-1)/x≥0
x(x²-2x-1)≥0
零点是1-√2,0,1+√2
则1-√2≤x≤0,x≥1+√2
分母不等于0
所以1-√2≤x<0,x≥1+√2
(x-1)/x>0
且2(x-1)/x)≥2
解方程组可得答案
log2(x-1/x)≥1 怎么算.急- 答案是x>1,x
log2^x怎么算
log2(x+1)≥0怎么解x的范围
|[log2(x)]^2-3log2(x)+1|
log2(x+1)>log2(3-x)
log2 (x+1)+log2 x=log6
log2(x+1)=3怎么解
f(-x)=log2 (-x+√x²+1)=log2[(x²+1-x²)/(x+√x²+1)]=-log2 (x+√x²+1)=-f(x) 其中的 log2[(x²+1-x²)/(x+√x²+1)]怎么算出-log2 (x+√x²+1)=-f(x) 的.我的数学工地很不好.
不等式log2 (x+1)
log2(2X-1)
log2(x²+1)
log2(x-1)
log2(x)=-1
log2(x+1)
log2(x-1)
log2(x-1)
log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1
log2(4*-x+1)怎么等于log2(4*x+1)-2x