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

[ 1/1Inequality]
    Assignment:Find the solution set of inequality （18000/(21X+1120))^2+(3000/(3500+7X))^2 <100^2 .
    Question type: Inequality
    Solution:
    The inequality can be reduced to 1 inequality：
         ( 18000 / ( 21 * X + 1120 ) ) ^ 2 + ( 3000 / ( 3500 + 7 * X ) ) ^ 2 <100 ^ 2         (1)
        From the definition field of divisor
         21 * x + 1120 ≠ 0        (2 )
        From the definition field of divisor
         3500 + 7 * x ≠ 0        (3 )

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    From inequality(1)：
         X < -61.905172 或  X ＞ -44.761525
    From inequality(2)：
         X < -160/3 或  X ＞ -160/3
    From inequality(3)：
         X < -500 或  X ＞ -500

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    From inequalities (1) and (2)
         X < -61.905172 或  X ＞ -44.761525    (4)
    From inequalities (3) and (4)
         X < -500 或  -500 < X < -61.905172 或  X ＞ -44.761525    (5)

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    The final solution set is :
         X < -500 或  -500 < X < -61.905172 或  X ＞ -44.761525

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