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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.
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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 k/(k*k+2*k) = {[(1-4*k*k)-1]/(k*k+1)-1}/5/{(k*k+1)-2*k} .
    Question type: Equation
    Solution:Original question:
      k ÷ ( k k + 2 k ) = (((1 − 4 k k ) − 1) ÷ ( k k + 1) − 1) ÷ 5 ÷ (( k k + 1) − 2 k )
     Multiply both sides of the equation by:( k k + 2 k ) ,  (( k k + 1) − 2 k )
      k (( k k + 1) − 2 k ) = (((1 − 4 k k ) − 1) ÷ ( k k + 1) − 1) ÷ 5 × ( k k + 2 k )
    Remove a bracket on the left of the equation:：
      k ( k k + 1) − k × 2 k = (((1 − 4 k k ) − 1) ÷ ( k k + 1) − 1) ÷ 5 × ( k k + 2 k )
    Remove a bracket on the right of the equation:：
      k ( k k + 1) − k × 2 k = ((1 − 4 k k ) − 1) ÷ ( k k + 1) ÷ 5 × ( k k + 2 k ) − 1 ÷ 5 × ( k k + 2 k )
    The equation is reduced to :
      k ( k k + 1) − k × 2 k = ((1 − 4 k k ) − 1) ÷ ( k k + 1) ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )
     Multiply both sides of the equation by:( k k + 1)
      k ( k k + 1)( k k + 1) − k × 2 k ( k k + 1) = ((1 − 4 k k ) − 1) ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the left of the equation:
      k k k ( k k + 1) + k × 1( k k + 1) − k × 2 k ( k k + 1) = ((1 − 4 k k ) − 1) ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the right of the equation:：
      k k k ( k k + 1) + k × 1( k k + 1) − k × 2 k ( k k + 1) = (1 − 4 k k ) ×
1
5
( k k + 2 k ) − 1 ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    The equation is reduced to :
      k k k ( k k + 1) + k × 1( k k + 1) − k × 2 k ( k k + 1) = (1 − 4 k k ) ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the left of the equation:
      k k k k k + k k k × 1 + k × 1( k k + 1) = (1 − 4 k k ) ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the right of the equation:：
      k k k k k + k k k × 1 + k × 1( k k + 1) = 1 ×
1
5
( k k + 2 k ) − 4 k k ×
1
5
( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1( k k + 1) =
1
5
( k k + 2 k ) −
4
5
 k k ( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the left of the equation:
      k k k k k + k k k × 1 + k × 1 k =
1
5
( k k + 2 k ) −
4
5
 k k ( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
( k k + 2 k )( k k + 1)
    Remove a bracket on the right of the equation:：
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
1
5
 × 2 k −
4
5
 k k ( k k + 2 k ) −
1
5
( k k + 2 k )
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k ( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
    Remove a bracket on the left of the equation:
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k ( k k + 2 k ) −
1
5
( k k + 2 k ) −
1
5
    Remove a bracket on the right of the equation:：
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k k k −
4
5
 k
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k k k −
8
5
 k
    Remove a bracket on the right of the equation:：
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k k k −
8
5
 k
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k +
2
5
 k −
4
5
 k k k k −
8
5
 k
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k + 0 k −
4
5
 k k k k −
8
5
 k
    Remove a bracket on the right of the equation:：
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k −
4
5
 k k k k −
8
5
 k k k
    The equation is reduced to :
      k k k k k + k k k × 1 + k × 1 k =
1
5
 k k −
4
5
 k k k k −
8
5
 k k k
    
    There are 0 solution(s).

解程的详细方法请参阅：《方程的解法》


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