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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-10)+1/(d-20) = 1/(d-5) .
    Question type: Equation
    Solution:Original question:
     2 ÷ d + 1 ÷ ( d − 10) + 1 ÷ ( d − 20) = 1 ÷ ( d − 5)
     Multiply both sides of the equation by: d  ,  ( d − 5)
     2( d − 5) + 1 ÷ ( d − 10) × d ( d − 5) + 1 ÷ ( d − 20) × d ( d − 5) = 1 d
    Remove a bracket on the left of the equation::
     2 d − 2 × 5 + 1 ÷ ( d − 10) × d ( d − 5) + 1 ÷ ( d − 20) × d ( d − 5) = 1 d
    The equation is reduced to :
     2 d − 10 + 1 ÷ ( d − 10) × d ( d − 5) + 1 ÷ ( d − 20) × d ( d − 5) = 1 d
     Multiply both sides of the equation by:( d − 10)
     2 d ( d − 10) − 10( d − 10) + 1 d ( d − 5) + 1 ÷ ( d − 20) × d ( d − 5) = 1 d ( d − 10)
    Remove a bracket on the left of the equation:
     2 d d − 2 d × 10 − 10( d − 10) + 1 d ( d − 5) + 1 = 1 d ( d − 10)
    Remove a bracket on the right of the equation::
     2 d d − 2 d × 10 − 10( d − 10) + 1 d ( d − 5) + 1 = 1 d d − 1 d × 10
    The equation is reduced to :
     2 d d − 20 d − 10( d − 10) + 1 d ( d − 5) + 1 ÷ ( d − 20) = 1 d d − 10 d
     Multiply both sides of the equation by:( d − 20)
     2 d d ( d − 20) − 20 d ( d − 20) − 10( d − 10)( d − 20) + 1 d = 1 d d ( d − 20) − 10 d ( d − 20)
    Remove a bracket on the left of the equation:
     2 d d d − 2 d d × 20 − 20 d ( d − 20) − 10 = 1 d d ( d − 20) − 10 d ( d − 20)
    Remove a bracket on the right of the equation::
     2 d d d − 2 d d × 20 − 20 d ( d − 20) − 10 = 1 d d d − 1 d d × 20 − 10 d ( d − 20)
    The equation is reduced to :
     2 d d d − 40 d d − 20 d ( d − 20) − 10( d − 10) = 1 d d d − 20 d d − 10 d ( d − 20)
    Remove a bracket on the left of the equation:
     2 d d d − 40 d d − 20 d d + 20 d = 1 d d d − 20 d d − 10 d ( d − 20)
    Remove a bracket on the right of the equation::
     2 d d d − 40 d d − 20 d d + 20 d = 1 d d d − 20 d d − 10 d d + 10 d
    The equation is reduced to :
     2 d d d − 40 d d − 20 d d + 400 d = 1 d d d − 20 d d − 10 d d + 200 d
    Remove a bracket on the left of the equation:
     2 d d d − 40 d d − 20 d d + 400 d = 1 d d d − 20 d d − 10 d d + 200 d
    The equation is reduced to :
     2 d d d − 40 d d − 20 d d + 400 d = 1 d d d − 20 d d − 10 d d + 200 d
    Remove a bracket on the left of the equation:
     2 d d d − 40 d d − 20 d d + 400 d = 1 d d d − 20 d d − 10 d d + 200 d
    The equation is reduced to :
     2 d d d − 40 d d − 20 d d + 400 d = 1 d d d − 20 d d − 10 d d + 200 d
    The equation is reduced to :
     2 d d d − 40 d d − 20 d d + 600 d = 1 d d d − 20 d d − 10 d d + 200 d
    Remove a bracket on the left of the equation:
     2 d d d − 40 d d − 20 d d + 600 d = 1 d d d − 20 d d − 10 d d + 200 d
    The equation is reduced to :
     2 d d d − 40 d d − 20 d d + 600 d = 1 d d d − 20 d d − 10 d d + 200 d

    
        d≈15.418685 , 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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