Overview: 1 questions will be solved this time.Among them
           ☆1 inequalities

[ 1/1Inequality]
    Assignment:Find the solution set of inequality 10^(-6)+10^(-21)*(f^2) >= (5*10^(-6)*f+500)/((2500*f+f+10^(8))*-0.15) .
    Question type: Inequality
    Solution:
    The inequality can be reduced to 1 inequality：
        10 ^ ( -6 ) + 10 ^ ( -21 ) * ( f ^ 2 ) >= ( 5 * 10 ^ ( -6 ) * f + 500 ) / ( ( 2500 * f + f + 10 ^ ( 8 ) ) * -0.15 )         (1)
        From the definition field of divisor
         ( 2500 * x + x + 10 ^ ( 8 ) ) * -0.15 ≠ 0        (2 )

---

    From inequality(1)：
         f ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!
    From inequality(2)：
         f ∈ R （R为全体实数），即在实数范围内，不等式恒成立！

---

    From inequalities (1) and (2)
         f ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!    (3)

---

    The final solution set is :
         f ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!

Your problem has not been solved here? Please take a look at the  hot problems ！