|1/(2x)+1|≥2
(x/x-1)+(5x-2)/x*x-x=1 求x(有图)
f(x)= x^2+1 x≥0
解不等式组 1-二分之一(x+1)≤x+2 ,x(x-1)≥(x+3)(x-3)
(x-1)^3(x+1)(x-2)/(x+4)
(x+1)(x+2)(x+3)(x+4)+1
计算 x(x+3)-(x-1)(x-2)
已知函数f(x)=x²+x+1,x≥0;2x+1,x
化简{x²-2x分之x+2-x²-4x+4x分之x-1}除以X分之(x-1)(X-4)乘以(2-x)
(x+2/x+1)-(x+4/x+3)-(x+3/x+2)+(x+5/x+4)
计算(x*x+4x+5)/(x+2)-(x*x+6x+10)/(x+3)+1
[(x+2)/(x²+2x)-(x-1)/(x²-4x+4)]÷(x-4)/x
化简 x(x-1)+2x(x+1)-3x(2x-5)