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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 2/d+1/(d+2)+1/(d+5) = 1/(d+0.5) .
    Question type: Equation
    Solution:Original question:
     2 ÷ d + 1 ÷ ( d + 2) + 1 ÷ ( d + 5) = 1 ÷ ( d +
1
2
)
     Multiply both sides of the equation by: d  ,  ( d +
1
2
)
     2( d +
1
2
) + 1 ÷ ( d + 2) × d ( d +
1
2
) + 1 ÷ ( d + 5) × d ( d +
1
2
) = 1 d
    Remove a bracket on the left of the equation:：
     2 d + 2 ×
1
2
 + 1 ÷ ( d + 2) × d ( d +
1
2
) + 1 ÷ ( d + 5) × d ( d +
1
2
) = 1 d
    The equation is reduced to :
     2 d + 1 + 1 ÷ ( d + 2) × d ( d +
1
2
) + 1 ÷ ( d + 5) × d ( d +
1
2
) = 1 d
     Multiply both sides of the equation by:( d + 2)
     2 d ( d + 2) + 1( d + 2) + 1 d ( d +
1
2
) + 1 ÷ ( d + 5) × d ( d +
1
2
) = 1 d ( d + 2)
    Remove a bracket on the left of the equation:
     2 d d + 2 d × 2 + 1( d + 2) + 1 d ( d +
1
2
) + 1 = 1 d ( d + 2)
    Remove a bracket on the right of the equation:：
     2 d d + 2 d × 2 + 1( d + 2) + 1 d ( d +
1
2
) + 1 = 1 d d + 1 d × 2
    The equation is reduced to :
     2 d d + 4 d + 1( d + 2) + 1 d ( d +
1
2
) + 1 ÷ ( d + 5) = 1 d d + 2 d
     Multiply both sides of the equation by:( d + 5)
     2 d d ( d + 5) + 4 d ( d + 5) + 1( d + 2)( d + 5) + 1 d = 1 d d ( d + 5) + 2 d ( d + 5)
    Remove a bracket on the left of the equation:
     2 d d d + 2 d d × 5 + 4 d ( d + 5) + 1 = 1 d d ( d + 5) + 2 d ( d + 5)
    Remove a bracket on the right of the equation:：
     2 d d d + 2 d d × 5 + 4 d ( d + 5) + 1 = 1 d d d + 1 d d × 5 + 2 d ( d + 5)
    The equation is reduced to :
     2 d d d + 10 d d + 4 d ( d + 5) + 1( d + 2) = 1 d d d + 5 d d + 2 d ( d + 5)
    Remove a bracket on the left of the equation:
     2 d d d + 10 d d + 4 d d + 4 d = 1 d d d + 5 d d + 2 d ( d + 5)
    Remove a bracket on the right of the equation:：
     2 d d d + 10 d d + 4 d d + 4 d = 1 d d d + 5 d d + 2 d d + 2 d
    The equation is reduced to :
     2 d d d + 10 d d + 4 d d + 20 d = 1 d d d + 5 d d + 2 d d + 10 d
    Remove a bracket on the left of the equation:
     2 d d d + 10 d d + 4 d d + 20 d = 1 d d d + 5 d d + 2 d d + 10 d
    The equation is reduced to :
     2 d d d + 10 d d + 4 d d + 20 d = 1 d d d + 5 d d + 2 d d + 10 d
    Remove a bracket on the left of the equation:
     2 d d d + 10 d d + 4 d d + 20 d = 1 d d d + 5 d d + 2 d d + 10 d
    The equation is reduced to :
     2 d d d + 10 d d + 4 d d + 20 d = 1 d d d + 5 d d + 2 d d + 10 d
    The equation is reduced to :
     2 d d d + 10 d d + 4 d d + 25 d = 1 d d d + 5 d d + 2 d d + 10 d
    Remove a bracket on the left of the equation:
     2 d d d + 10 d d + 4 d d + 25 d = 1 d d d + 5 d d + 2 d d + 10 d
    The equation is reduced to :
     2 d d d + 10 d d + 4 d d + 25 d = 1 d d d + 5 d d + 2 d d + 10 d

    
        d≈-3.747108 , 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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