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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+2) = 1/(d+5) .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( d + 3) + 1 ÷ ( d + 2) = 1 ÷ ( d + 5)
     Multiply both sides of the equation by:( d + 3) ,  ( d + 5)
     1( d + 5) + 1 ÷ ( d + 2) × ( d + 3)( d + 5) = 1( d + 3)
    Remove a bracket on the left of the equation::
     1 d + 1 × 5 + 1 ÷ ( d + 2) × ( d + 3)( d + 5) = 1( d + 3)
    Remove a bracket on the right of the equation::
     1 d + 1 × 5 + 1 ÷ ( d + 2) × ( d + 3)( d + 5) = 1 d + 1 × 3
    The equation is reduced to :
     1 d + 5 + 1 ÷ ( d + 2) × ( d + 3)( d + 5) = 1 d + 3
     Multiply both sides of the equation by:( d + 2)
     1 d ( d + 2) + 5( d + 2) + 1( d + 3)( d + 5) = 1 d ( d + 2) + 3( d + 2)
    Remove a bracket on the left of the equation:
     1 d d + 1 d × 2 + 5( d + 2) + 1( d + 3)( d + 5) = 1 d ( d + 2) + 3( d + 2)
    Remove a bracket on the right of the equation::
     1 d d + 1 d × 2 + 5( d + 2) + 1( d + 3)( d + 5) = 1 d d + 1 d × 2 + 3( d + 2)
    The equation is reduced to :
     1 d d + 2 d + 5( d + 2) + 1( d + 3)( d + 5) = 1 d d + 2 d + 3( d + 2)
    Remove a bracket on the left of the equation:
     1 d d + 2 d + 5 d + 5 × 2 + 1( d + 3)( d + 5) = 1 d d + 2 d + 3( d + 2)
    Remove a bracket on the right of the equation::
     1 d d + 2 d + 5 d + 5 × 2 + 1( d + 3)( d + 5) = 1 d d + 2 d + 3 d + 3 × 2
    The equation is reduced to :
     1 d d + 2 d + 5 d + 10 + 1( d + 3)( d + 5) = 1 d d + 2 d + 3 d + 6
    The equation is reduced to :
     1 d d + 7 d + 10 + 1( d + 3)( d + 5) = 1 d d + 5 d + 6
    Remove a bracket on the left of the equation:
     1 d d + 7 d + 10 + 1 d ( d + 5) + 1 × 3( d + 5) = 1 d d + 5 d + 6
    The equation is reduced to :
     1 d d + 7 d + 10 + 1 d ( d + 5) + 3( d + 5) = 1 d d + 5 d + 6
    Remove a bracket on the left of the equation:
     1 d d + 7 d + 10 + 1 d d + 1 d × 5 = 1 d d + 5 d + 6
    The equation is reduced to :
     1 d d + 7 d + 10 + 1 d d + 5 d + 3 = 1 d d + 5 d + 6
    The equation is reduced to :
     1 d d + 12 d + 10 + 1 d d + 3( d + 5) = 1 d d + 5 d + 6
    Remove a bracket on the left of the equation:
     1 d d + 12 d + 10 + 1 d d + 3 d + 3 = 1 d d + 5 d + 6
    The equation is reduced to :
     1 d d + 12 d + 10 + 1 d d + 3 d + 15 = 1 d d + 5 d + 6
    The equation is reduced to :
     1 d d + 15 d + 25 + 1 d d = 1 d d + 5 d + 6

    After the equation is converted into a general formula, there is a common factor:
    ( √3500000d +√194232141 )
    From
        √3500000d +√194232141 = 0

    it is concluded that::
        d1=-
√194232141
√3500000

    Solutions that cannot be obtained by factorization:
        d2≈-2.550510 , 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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