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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+3)+1/(d+5)+1/d = 1/(d+7) .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( d + 3) + 1 ÷ ( d + 5) + 1 ÷ d = 1 ÷ ( d + 7)
     Multiply both sides of the equation by:( d + 3) ,  ( d + 7)
     1( d + 7) + 1 ÷ ( d + 5) × ( d + 3)( d + 7) + 1 ÷ d × ( d + 3)( d + 7) = 1( d + 3)
    Remove a bracket on the left of the equation::
     1 d + 1 × 7 + 1 ÷ ( d + 5) × ( d + 3)( d + 7) + 1 ÷ d × ( d + 3)( d + 7) = 1( d + 3)
    Remove a bracket on the right of the equation::
     1 d + 1 × 7 + 1 ÷ ( d + 5) × ( d + 3)( d + 7) + 1 ÷ d × ( d + 3)( d + 7) = 1 d + 1 × 3
    The equation is reduced to :
     1 d + 7 + 1 ÷ ( d + 5) × ( d + 3)( d + 7) + 1 ÷ d × ( d + 3)( d + 7) = 1 d + 3
     Multiply both sides of the equation by:( d + 5)
     1 d ( d + 5) + 7( d + 5) + 1( d + 3)( d + 7) + 1 ÷ d × ( d + 3)( d + 7) = 1 d ( d + 5) + 3( d + 5)
    Remove a bracket on the left of the equation:
     1 d d + 1 d × 5 + 7( d + 5) + 1( d + 3)( d + 7) + 1 = 1 d ( d + 5) + 3( d + 5)
    Remove a bracket on the right of the equation::
     1 d d + 1 d × 5 + 7( d + 5) + 1( d + 3)( d + 7) + 1 = 1 d d + 1 d × 5 + 3( d + 5)
    The equation is reduced to :
     1 d d + 5 d + 7( d + 5) + 1( d + 3)( d + 7) + 1 ÷ d = 1 d d + 5 d + 3( d + 5)
     Multiply both sides of the equation by: d
     1 d d d + 5 d d + 7( d + 5) d + 1( d + 3) = 1 d d d + 5 d d + 3( d + 5) d
    Remove a bracket on the left of the equation:
     1 d d d + 5 d d + 7 d d + 7 × 5 = 1 d d d + 5 d d + 3( d + 5) d
    Remove a bracket on the right of the equation::
     1 d d d + 5 d d + 7 d d + 7 × 5 = 1 d d d + 5 d d + 3 d d + 3 × 5
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    Remove a bracket on the left of the equation:
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    Remove a bracket on the left of the equation:
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    Remove a bracket on the left of the equation:
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 35 d = 1 d d d + 5 d d + 3 d d + 15 d
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 56 d = 1 d d d + 5 d d + 3 d d + 15 d
    Remove a bracket on the left of the equation:
     1 d d d + 5 d d + 7 d d + 56 d = 1 d d d + 5 d d + 3 d d + 15 d
    The equation is reduced to :
     1 d d d + 5 d d + 7 d d + 56 d = 1 d d d + 5 d d + 3 d d + 15 d

    The solution of the equation:
        d1≈-8.833060 , keep 6 decimal places
        d2≈-4.277413 , keep 6 decimal places
        d3≈-1.389527 , keep 6 decimal places
    
    There are 3 solution(s).


解程的详细方法请参阅:《方程的解法》



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