Mathematics
         
语言:中文    Language:English
                                Equations   
Fold
                                Unary equation
                                Multivariate equation
                                Math OP  
Unfold
                                Linear algebra      
Unfold
                                Derivative function
                                Function image
                                Hot issues
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+3) = 1/(D+1)+1/(D+2) .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( D + 3) = 1 ÷ ( D + 1) + 1 ÷ ( D + 2)
     Multiply both sides of the equation by:( D + 3) ,  ( D + 1)
     1( D + 1) = 1( D + 3) + 1 ÷ ( D + 2) × ( D + 3)( D + 1)
    Remove a bracket on the left of the equation::
     1 D + 1 × 1 = 1( D + 3) + 1 ÷ ( D + 2) × ( D + 3)( D + 1)
    Remove a bracket on the right of the equation::
     1 D + 1 × 1 = 1 D + 1 × 3 + 1 ÷ ( D + 2) × ( D + 3)( D + 1)
    The equation is reduced to :
     1 D + 1 = 1 D + 3 + 1 ÷ ( D + 2) × ( D + 3)( D + 1)
     Multiply both sides of the equation by:( D + 2)
     1 D ( D + 2) + 1( D + 2) = 1 D ( D + 2) + 3( D + 2) + 1( D + 3)( D + 1)
    Remove a bracket on the left of the equation:
     1 D D + 1 D × 2 + 1( D + 2) = 1 D ( D + 2) + 3( D + 2) + 1( D + 3)( D + 1)
    Remove a bracket on the right of the equation::
     1 D D + 1 D × 2 + 1( D + 2) = 1 D D + 1 D × 2 + 3( D + 2) + 1( D + 3)( D + 1)
    The equation is reduced to :
     1 D D + 2 D + 1( D + 2) = 1 D D + 2 D + 3( D + 2) + 1( D + 3)( D + 1)
    Remove a bracket on the left of the equation:
     1 D D + 2 D + 1 D + 1 × 2 = 1 D D + 2 D + 3( D + 2) + 1( D + 3)( D + 1)
    Remove a bracket on the right of the equation::
     1 D D + 2 D + 1 D + 1 × 2 = 1 D D + 2 D + 3 D + 3 × 2 + 1( D + 3)( D + 1)
    The equation is reduced to :
     1 D D + 2 D + 1 D + 2 = 1 D D + 2 D + 3 D + 6 + 1( D + 3)( D + 1)
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 5 D + 6 + 1( D + 3)( D + 1)
    Remove a bracket on the right of the equation::
     1 D D + 3 D + 2 = 1 D D + 5 D + 6 + 1 D ( D + 1) + 1 × 3( D + 1)
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 5 D + 6 + 1 D ( D + 1) + 3( D + 1)
    Remove a bracket on the right of the equation::
     1 D D + 3 D + 2 = 1 D D + 5 D + 6 + 1 D D + 1 D × 1
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 5 D + 6 + 1 D D + 1 D + 3
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 6 D + 6 + 1 D D + 3( D + 1)
    Remove a bracket on the right of the equation::
     1 D D + 3 D + 2 = 1 D D + 6 D + 6 + 1 D D + 3 D + 3
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 6 D + 6 + 1 D D + 3 D + 3
    The equation is reduced to :
     1 D D + 3 D + 2 = 1 D D + 9 D + 9 + 1 D D

    The solution of the equation:
        D1≈-4.414214 , keep 6 decimal places
        D2≈-1.585786 , keep 6 decimal places
    
    There are 2 solution(s).


解程的详细方法请参阅:《方程的解法》



Your problem has not been solved here? Please go to the Hot Problems section!





  New addition:Lenders ToolBox module(Specific location:Math OP > Lenders ToolBox ),welcome。