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

    The solution of the equation:
        d1≈-2.366025 , keep 6 decimal places
        d2≈-0.633975 , 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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