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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+1)+1/(d+4) = 1/(d+2) .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( d ) + 1 ÷ ( d + 1) + 1 ÷ ( d + 4) = 1 ÷ ( d + 2)
     Multiply both sides of the equation by:( d ) ,  ( d + 2)
     1( d + 2) + 1 ÷ ( d + 1) × ( d )( d + 2) + 1 ÷ ( d + 4) × ( d )( d + 2) = 1( d )
    Remove a bracket on the left of the equation::
     1 d + 1 × 2 + 1 ÷ ( d + 1) × ( d )( d + 2) + 1 ÷ ( d + 4) × ( d )( d + 2) = 1( d )
    Remove a bracket on the right of the equation::
     1 d + 1 × 2 + 1 ÷ ( d + 1) × ( d )( d + 2) + 1 ÷ ( d + 4) × ( d )( d + 2) = 1 d
    The equation is reduced to :
     1 d + 2 + 1 ÷ ( d + 1) × ( d )( d + 2) + 1 ÷ ( d + 4) × ( d )( d + 2) = 1 d
     Multiply both sides of the equation by:( d + 1)
     1 d ( d + 1) + 2( d + 1) + 1( d )( d + 2) + 1 ÷ ( d + 4) × ( d )( d + 2) = 1 d ( d + 1)
    Remove a bracket on the left of the equation:
     1 d d + 1 d × 1 + 2( d + 1) + 1( d )( d + 2) + 1 = 1 d ( d + 1)
    Remove a bracket on the right of the equation::
     1 d d + 1 d × 1 + 2( d + 1) + 1( d )( d + 2) + 1 = 1 d d + 1 d × 1
    The equation is reduced to :
     1 d d + 1 d + 2( d + 1) + 1( d )( d + 2) + 1 ÷ ( d + 4) = 1 d d + 1 d
     Multiply both sides of the equation by:( d + 4)
     1 d d ( d + 4) + 1 d ( d + 4) + 2( d + 1)( d + 4) + 1( d ) = 1 d d ( d + 4) + 1 d ( d + 4)
    Remove a bracket on the left of the equation:
     1 d d d + 1 d d × 4 + 1 d ( d + 4) + 2 = 1 d d ( d + 4) + 1 d ( d + 4)
    Remove a bracket on the right of the equation::
     1 d d d + 1 d d × 4 + 1 d ( d + 4) + 2 = 1 d d d + 1 d d × 4 + 1 d ( d + 4)
    The equation is reduced to :
     1 d d d + 4 d d + 1 d ( d + 4) + 2( d + 1) = 1 d d d + 4 d d + 1 d ( d + 4)
    Remove a bracket on the left of the equation:
     1 d d d + 4 d d + 1 d d + 1 d = 1 d d d + 4 d d + 1 d ( d + 4)
    Remove a bracket on the right of the equation::
     1 d d d + 4 d d + 1 d d + 1 d = 1 d d d + 4 d d + 1 d d + 1 d
    The equation is reduced to :
     1 d d d + 4 d d + 1 d d + 4 d = 1 d d d + 4 d d + 1 d d + 4 d
    Remove a bracket on the left of the equation:
     1 d d d + 4 d d + 1 d d + 4 d = 1 d d d + 4 d d + 1 d d + 4 d
    The equation is reduced to :
     1 d d d + 4 d d + 1 d d + 4 d = 1 d d d + 4 d d + 1 d d + 4 d
    Remove a bracket on the left of the equation:
     1 d d d + 4 d d + 1 d d + 4 d = 1 d d d + 4 d d + 1 d d + 4 d
    The equation is reduced to :
     1 d d d + 4 d d + 1 d d + 4 d = 1 d d d + 4 d d + 1 d d + 4 d
    The equation is reduced to :
     1 d d d + 4 d d + 1 d d + 12 d = 1 d d d + 4 d d + 1 d d + 4 d
    Remove a bracket on the left of the equation:
     1 d d d + 4 d d + 1 d d + 12 d = 1 d d d + 4 d d + 1 d d + 4 d

    
        d≈-0.549460 , keep 6 decimal places
    
    There are 1 solution(s).


解一元一次方程的详细方法请参阅:《一元一次方程的解法》



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