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On line Solution of Monovariate Equation:
    Input any unary equation directly, and then click the "Next" button to obtain the solution of the equation.
    It supports equations that contain mathematical functions.
    Current location:Equations > Monovariate Equation > The history of univariate equation calculation > Answer

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

[ 1/1 Equation]
    Work: Find the solution of equation (1/d)+[1/(d+5)]+[1/(d+10)] = 0 .
    Question type: Equation
    Solution:Original question:
     (1 ÷ d ) + (1 ÷ ( d + 5)) + (1 ÷ ( d + 10)) = 0
    Remove a bracket on the left of the equation::
     1 ÷ d + (1 ÷ ( d + 5)) + (1 ÷ ( d + 10)) = 0
     Multiply both sides of the equation by: d
     1 + (1 ÷ ( d + 5)) d + (1 ÷ ( d + 10)) d = 0
    Remove a bracket on the left of the equation:
     1 + 1 ÷ ( d + 5) × d + (1 ÷ ( d + 10)) d = 0
     Multiply both sides of the equation by:( d + 5)
     1( d + 5) + 1 d + (1 ÷ ( d + 10)) d ( d + 5) = 0
    Remove a bracket on the left of the equation:
     1 d + 1 × 5 + 1 d + (1 ÷ ( d + 10)) d ( d + 5) = 0
    The equation is reduced to :
     1 d + 5 + 1 d + (1 ÷ ( d + 10)) d ( d + 5) = 0
    The equation is reduced to :
     2 d + 5 + (1 ÷ ( d + 10)) d ( d + 5) = 0
    Remove a bracket on the left of the equation:
     2 d + 5 + 1 ÷ ( d + 10) × d ( d + 5) = 0
     Multiply both sides of the equation by:( d + 10)
     2 d ( d + 10) + 5( d + 10) + 1 d ( d + 5) = 0
    Remove a bracket on the left of the equation:
     2 d d + 2 d × 10 + 5( d + 10) + 1 d ( d + 5) = 0
    The equation is reduced to :
     2 d d + 20 d + 5( d + 10) + 1 d ( d + 5) = 0
    Remove a bracket on the left of the equation:
     2 d d + 20 d + 5 d + 5 × 10 + 1 d ( d + 5) = 0
    The equation is reduced to :
     2 d d + 20 d + 5 d + 50 + 1 d ( d + 5) = 0
    The equation is reduced to :
     2 d d + 25 d + 50 + 1 d ( d + 5) = 0
    Remove a bracket on the left of the equation:
     2 d d + 25 d + 50 + 1 d d + 1 d × 5 = 0
    The equation is reduced to :
     2 d d + 25 d + 50 + 1 d d + 5 d = 0
    The equation is reduced to :
     2 d d + 30 d + 50 + 1 d d = 0

    The solution of the equation:
        d1≈-7.886751 , keep 6 decimal places
        d2≈-2.113249 , keep 6 decimal places
    
    There are 2 solution(s).


解程的详细方法请参阅:《方程的解法》



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