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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/（x+0.1） = 1/（x-0.5）+1/（x-5） .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( x +
1
10
) = 1 ÷ ( x −
1
2
) + 1 ÷ ( x − 5)
     Multiply both sides of the equation by:( x +
1
10
) ,  ( x −
1
2
)
     1( x −
1
2
) = 1( x +
1
10
) + 1 ÷ ( x − 5) × ( x +
1
10
)( x −
1
2
)
    Remove a bracket on the left of the equation:：
     1 x − 1 ×
1
2
 = 1( x +
1
10
) + 1 ÷ ( x − 5) × ( x +
1
10
)( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x − 1 ×
1
2
 = 1 x + 1 ×
1
10
 + 1 ÷ ( x − 5) × ( x +
1
10
)( x −
1
2
)
    The equation is reduced to :
     1 x −
1
2
 = 1 x +
1
10
 + 1 ÷ ( x − 5) × ( x +
1
10
)( x −
1
2
)
     Multiply both sides of the equation by:( x − 5)
     1 x ( x − 5) −
1
2
( x − 5) = 1 x ( x − 5) +
1
10
( x − 5) + 1( x +
1
10
)( x −
1
2
)
    Remove a bracket on the left of the equation:
     1 x x − 1 x × 5 −
1
2
( x − 5) = 1 x ( x − 5) +
1
10
( x − 5) + 1( x +
1
10
)( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x x − 1 x × 5 −
1
2
( x − 5) = 1 x x − 1 x × 5 +
1
10
( x − 5) + 1( x +
1
10
)( x −
1
2
)
    The equation is reduced to :
     1 x x − 5 x −
1
2
( x − 5) = 1 x x − 5 x +
1
10
( x − 5) + 1( x +
1
10
)( x −
1
2
)
    Remove a bracket on the left of the equation:
     1 x x − 5 x −
1
2
 x +
1
2
 × 5 = 1 x x − 5 x +
1
10
( x − 5) + 1( x +
1
10
)( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x x − 5 x −
1
2
 x +
1
2
 × 5 = 1 x x − 5 x +
1
10
 x −
1
10
 × 5 + 1( x +
1
10
)( x −
1
2
)
    The equation is reduced to :
     1 x x − 5 x −
1
2
 x +
5
2
 = 1 x x − 5 x +
1
10
 x −
1
2
 + 1( x +
1
10
)( x −
1
2
)
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
49
10
 x −
1
2
 + 1( x +
1
10
)( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x x −
11
2
 x +
5
2
 = 1 x x −
49
10
 x −
1
2
 + 1 x ( x −
1
2
) + 1 ×
1
10
( x −
1
2
)
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
49
10
 x −
1
2
 + 1 x ( x −
1
2
) +
1
10
( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x x −
11
2
 x +
5
2
 = 1 x x −
49
10
 x −
1
2
 + 1 x x − 1 x ×
1
2
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
49
10
 x −
1
2
 + 1 x x −
1
2
 x +
1
10
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
27
5
 x −
1
2
 + 1 x x +
1
10
( x −
1
2
)
    Remove a bracket on the right of the equation:：
     1 x x −
11
2
 x +
5
2
 = 1 x x −
27
5
 x −
1
2
 + 1 x x +
1
10
 x −
1
10
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
27
5
 x −
1
2
 + 1 x x +
1
10
 x −
1
20
    The equation is reduced to :
     1 x x −
11
2
 x +
5
2
 = 1 x x −
53
10
 x −
11
20
 + 1 x x

    After the equation is converted into a general formula, it is converted into:
    ( x + 1.849286 )( x - 1.649286 )=0
    From
        x + 1.849286 = 0
        x - 1.649286 = 0
    it is concluded that:：
        x1≈-1.849286 , keep 6 decimal places
        x1≈1.649286 , keep 6 decimal places
    Solutions that cannot be obtained by factorization:
        x2=    
    There are 2 solution(s).

解一元二次方程的详细方法请参阅：《一元二次方程的解法》


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