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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+5) = 1/(d+2)+1/(d+3)+1/d .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( d + 5) = 1 ÷ ( d + 2) + 1 ÷ ( d + 3) + 1 ÷ d
     Multiply both sides of the equation by:( d + 5) ,  ( d + 2)
     1( d + 2) = 1( d + 5) + 1 ÷ ( d + 3) × ( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
    Remove a bracket on the left of the equation::
     1 d + 1 × 2 = 1( d + 5) + 1 ÷ ( d + 3) × ( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
    Remove a bracket on the right of the equation::
     1 d + 1 × 2 = 1 d + 1 × 5 + 1 ÷ ( d + 3) × ( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
    The equation is reduced to :
     1 d + 2 = 1 d + 5 + 1 ÷ ( d + 3) × ( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
     Multiply both sides of the equation by:( d + 3)
     1 d ( d + 3) + 2( d + 3) = 1 d ( d + 3) + 5( d + 3) + 1( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
    Remove a bracket on the left of the equation:
     1 d d + 1 d × 3 + 2( d + 3) = 1 d ( d + 3) + 5( d + 3) + 1( d + 5)( d + 2) + 1 ÷ d × ( d + 5)( d + 2)
    Remove a bracket on the right of the equation::
     1 d d + 1 d × 3 + 2( d + 3) = 1 d d + 1 d × 3 + 5( d + 3) + 1( d + 5)( d + 2) + 1
    The equation is reduced to :
     1 d d + 3 d + 2( d + 3) = 1 d d + 3 d + 5( d + 3) + 1( d + 5)( d + 2) + 1 ÷ d
     Multiply both sides of the equation by: d
     1 d d d + 3 d d + 2( d + 3) d = 1 d d d + 3 d d + 5( d + 3) d + 1( d + 5)
    Remove a bracket on the left of the equation:
     1 d d d + 3 d d + 2 d d + 2 × 3 = 1 d d d + 3 d d + 5( d + 3) d + 1( d + 5)
    Remove a bracket on the right of the equation::
     1 d d d + 3 d d + 2 d d + 2 × 3 = 1 d d d + 3 d d + 5 d d + 5 × 3
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    Remove a bracket on the right of the equation::
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    Remove a bracket on the right of the equation::
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    Remove a bracket on the right of the equation::
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 15 d
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 25 d
    Remove a bracket on the right of the equation::
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 25 d
    The equation is reduced to :
     1 d d d + 3 d d + 2 d d + 6 d = 1 d d d + 3 d d + 5 d d + 25 d

    The solution of the equation:
        d1≈-6.517113 , keep 6 decimal places
        d2≈-2.596424 , keep 6 decimal places
        d3≈-0.886462 , 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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