6x^2-640x+634≥0
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/13 03:38:44
6x^2-640x+634≥06x^2-640x+634≥06x^2-640x+634≥06x^2-640x+634=(2x-2)(3x-317)>=0所以x>=317/3或x以开发个igkni(纯属
6x^2-640x+634≥0
6x^2-640x+634≥0
6x^2-640x+634≥0
6x^2-640x+634
=(2x-2)(3x-317)>=0
所以x>=317/3或x
以开发个igkni (纯属虚构)
6x²-640x+634≥0
3x²-320x+317≥0
(3x-317)(x-1)≥0
即x≥317/3或x≤1
6x^2-640x+634≥0
3≤|8-x| |x+1|>2-x |x+3|7|x+1| x^2-6|x|+9>0 |x-1|+|x+2|≥5 (1)方法
f(x)分段函数-x^2+x(x≥0),x^2+x(x
设偶函数f(x)满足f(x)=2ˆx-4(x≥0),则{x|f(x-2)>0}=1.{x|x〈-2或x〉4}2.{x|x〈0或X>4}3.{x|x<0或x>6}4.{x|x<-2或x>2}
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
x(x*x-6x-9)-x(x*x-8x-15)+2x(3-x)
填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x
x(x+2)(x+6)+20x+6=0
(x^-2x)^-5(x^-2x)-6=0
(x+2)(x-3)(x-6)(x-1)=0
6/(x-1)-(x+2)/x(x-1)=0
已知1+x+x^2+x^3=0,求x+x^2+x^3+x^4+x^+x^6+x^7+x^8的值
如果1+x+x^2+x^3=0,求x+x^2+X^3+x^4+x^5+x^6+x^7+x^8的值,
1/(x*x-2x-3)+2/(x*-x-6)+3/(x*x+3x+2)=0
y=x^2-2x (x≥0) -x^2-2x(x
求y=(x^2)+2x(x≥0) y=-x+2x(x
已知函数f(x)=x²+x+1,x≥0;2x+1,x
已知f(x)={x²+x+1,x≥0 2x+1,x