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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+2)+1/(x-1/2)-2/(x-2) = 0 .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( x + 2) + 1 ÷ ( x 1 ÷ 2)2 ÷ ( x 2) = 0
     Multiply both sides of the equation by:( x + 2)
     1 + 1 ÷ ( x 1 ÷ 2) × ( x + 2)2 ÷ ( x 2) × ( x + 2) = 0
    Remove a bracket on the left of the equation::
     1 + 1 ÷ ( x 1 ÷ 2) × x + 1 ÷ ( x 1 ÷ 2) × 22 ÷ ( x 2) × ( x + 2) = 0
    The equation is reduced to :
     1 + 1 ÷ ( x 1 ÷ 2) × x + 2 ÷ ( x 1 ÷ 2)2 ÷ ( x 2) × ( x + 2) = 0
     Multiply both sides of the equation by:( x 1 ÷ 2)
     1( x 1 ÷ 2) + 1 x + 22 ÷ ( x 2) × ( x + 2)( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     1 x 1 × 1 ÷ 2 + 1 x + 22 ÷ ( x 2) × ( x + 2)( x 1 ÷ 2) = 0
    The equation is reduced to :
     1 x
1
2
+ 1 x + 22 ÷ ( x 2) × ( x + 2)( x 1 ÷ 2) = 0
    The equation is reduced to :
     2 x +
3
2
2 ÷ ( x 2) × ( x + 2)( x 1 ÷ 2) = 0
     Multiply both sides of the equation by:( x 2)
     2 x ( x 2) +
3
2
( x 2)2( x + 2)( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     2 x x 2 x × 2 +
3
2
( x 2)2( x + 2)( x 1 ÷ 2) = 0
    The equation is reduced to :
     2 x x 4 x +
3
2
( x 2)2( x + 2)( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     2 x x 4 x +
3
2
x
3
2
× 22( x + 2)( x 1 ÷ 2) = 0
    The equation is reduced to :
     2 x x 4 x +
3
2
x 32( x + 2)( x 1 ÷ 2) = 0
    The equation is reduced to :
     2 x x
5
2
x 32( x + 2)( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     2 x x
5
2
x 32 x ( x 1 ÷ 2)2 × 2( x 1 ÷ 2) = 0
    The equation is reduced to :
     2 x x
5
2
x 32 x ( x 1 ÷ 2)4( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     2 x x
5
2
x 32 x x + 2 x × 1 = 0
    The equation is reduced to :
     2 x x
5
2
x 32 x x + 1 x 4 = 0
    The equation is reduced to :
     2 x x
3
2
x 32 x x 4( x 1 ÷ 2) = 0
    Remove a bracket on the left of the equation:
     2 x x
3
2
x 32 x x 4 x + 4 = 0
    The equation is reduced to :
     2 x x
3
2
x 32 x x 4 x + 2 = 0
    The equation is reduced to :
     2 x x
11
2
x 12 x x = 0

    
        x1=-
2
11
    
    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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