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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+1.5) .
    Question type: Equation
    Solution:Original question:
     1 ÷ d + 1 ÷ ( d + 1) + 1 ÷ ( d + 4) = 1 ÷ ( d +
3
2
)
     Multiply both sides of the equation by: d  ,  ( d +
3
2
)
     1( d +
3
2
) + 1 ÷ ( d + 1) × d ( d +
3
2
) + 1 ÷ ( d + 4) × d ( d +
3
2
) = 1 d
    Remove a bracket on the left of the equation::
     1 d + 1 ×
3
2
+ 1 ÷ ( d + 1) × d ( d +
3
2
) + 1 ÷ ( d + 4) × d ( d +
3
2
) = 1 d
    The equation is reduced to :
     1 d +
3
2
+ 1 ÷ ( d + 1) × d ( d +
3
2
) + 1 ÷ ( d + 4) × d ( d +
3
2
) = 1 d
     Multiply both sides of the equation by:( d + 1)
     1 d ( d + 1) +
3
2
( d + 1) + 1 d ( d +
3
2
) + 1 ÷ ( d + 4) × d ( d +
3
2
) = 1 d ( d + 1)
    Remove a bracket on the left of the equation:
     1 d d + 1 d × 1 +
3
2
( d + 1) + 1 d ( d +
3
2
) + 1 = 1 d ( d + 1)
    Remove a bracket on the right of the equation::
     1 d d + 1 d × 1 +
3
2
( d + 1) + 1 d ( d +
3
2
) + 1 = 1 d d + 1 d × 1
    The equation is reduced to :
     1 d d + 1 d +
3
2
( d + 1) + 1 d ( d +
3
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) +
3
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) +
3
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) +
3
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) +
3
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 + 10 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 + 10 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 + 10 d = 1 d d d + 4 d d + 1 d d + 4 d

    
        d≈-0.597921 , 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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