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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+2)+1/(d-0.5) = 2/(d-2) .
    Question type: Equation
    Solution:Original question:
     1 ÷ ( d + 2) + 1 ÷ ( d
1
2
) = 2 ÷ ( d 2)
     Multiply both sides of the equation by:( d + 2) ,  ( d 2)
     1( d 2) + 1 ÷ ( d
1
2
) × ( d + 2)( d 2) = 2( d + 2)
    Remove a bracket on the left of the equation::
     1 d 1 × 2 + 1 ÷ ( d
1
2
) × ( d + 2)( d 2) = 2( d + 2)
    Remove a bracket on the right of the equation::
     1 d 1 × 2 + 1 ÷ ( d
1
2
) × ( d + 2)( d 2) = 2 d + 2 × 2
    The equation is reduced to :
     1 d 2 + 1 ÷ ( d
1
2
) × ( d + 2)( d 2) = 2 d + 4
     Multiply both sides of the equation by:( d
1
2
)
     1 d ( d
1
2
)2( d
1
2
) + 1( d + 2)( d 2) = 2 d ( d
1
2
) + 4( d
1
2
)
    Remove a bracket on the left of the equation:
     1 d d 1 d ×
1
2
2( d
1
2
) + 1( d + 2)( d 2) = 2 d ( d
1
2
) + 4( d
1
2
)
    Remove a bracket on the right of the equation::
     1 d d 1 d ×
1
2
2( d
1
2
) + 1( d + 2)( d 2) = 2 d d 2 d ×
1
2
+ 4( d
1
2
)
    The equation is reduced to :
     1 d d
1
2
d 2( d
1
2
) + 1( d + 2)( d 2) = 2 d d 1 d + 4( d
1
2
)
    Remove a bracket on the left of the equation:
     1 d d
1
2
d 2 d + 2 ×
1
2
+ 1( d + 2)( d 2) = 2 d d 1 d + 4( d
1
2
)
    Remove a bracket on the right of the equation::
     1 d d
1
2
d 2 d + 2 ×
1
2
+ 1( d + 2)( d 2) = 2 d d 1 d + 4 d 4 ×
1
2
    The equation is reduced to :
     1 d d
1
2
d 2 d + 1 + 1( d + 2)( d 2) = 2 d d 1 d + 4 d 2
    The equation is reduced to :
     1 d d
5
2
d + 1 + 1( d + 2)( d 2) = 2 d d + 3 d 2
    Remove a bracket on the left of the equation:
     1 d d
5
2
d + 1 + 1 d ( d 2) + 1 × 2( d 2) = 2 d d + 3 d 2
    The equation is reduced to :
     1 d d
5
2
d + 1 + 1 d ( d 2) + 2( d 2) = 2 d d + 3 d 2
    Remove a bracket on the left of the equation:
     1 d d
5
2
d + 1 + 1 d d 1 d × 2 = 2 d d + 3 d 2
    The equation is reduced to :
     1 d d
5
2
d + 1 + 1 d d 2 d + 2 = 2 d d + 3 d 2
    The equation is reduced to :
     1 d d
9
2
d + 1 + 1 d d + 2( d 2) = 2 d d + 3 d 2
    Remove a bracket on the left of the equation:
     1 d d
9
2
d + 1 + 1 d d + 2 d 2 = 2 d d + 3 d 2
    The equation is reduced to :
     1 d d
9
2
d + 1 + 1 d d + 2 d 4 = 2 d d + 3 d 2
    The equation is reduced to :
     1 d d
5
2
d 3 + 1 d d = 2 d d + 3 d 2

    
        d1=-
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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