Respuesta :
The way how the question entered here, I have to make few assumptions to answer. Hoping that one of the below assumption is your actual problem.
Assumption 1: If your complex fraction is given as below
[tex] \frac{\frac{-2}{x+\frac{5}{y}}}{\frac{3}{y-\frac{2}{x}}} [/tex]
Then we need to write [tex] x+\frac{5}{y} [/tex] as a single fraction by rewriting x as a fraction [tex] \frac{x}{1} [/tex], then multiplying y on both top and bottom of that fraction to get [tex] \frac{xy}{y} [/tex].
We can rewrite [tex] x+\frac{5}{y} [/tex] as [tex] \frac{xy}{y} +\frac{5}{y} [/tex] which will be equal to [tex] \frac{xy+5}{y} [/tex].
In the same way[tex] y-\frac{2}{x} =\frac{xy}{x}-\frac{2}{x}=\frac{xy-2}{x} [/tex]
When dividing fractions we need to flip the bottom and multiply with top:
So [tex] \frac{-2}{\frac{xy+5}{y}} [/tex][tex] \frac{-2}{1} * \frac{y}{xy+5}=\frac{-2y}{xy+5} [/tex]
In the same way, [tex] \frac{3}{\frac{xy-2}{x}} =\frac{3}{1} *\frac{x}{xy-2} =\frac{3x}{xy-2} [/tex]
Applying the same logic of flipping the fraction and multiplying with the top, we get the below final expression.
[tex] \frac{\frac{-2y}{xy+5} }{\frac{3x}{xy-2} } =\frac{-2y}{xy+5}*\frac{xy-2}{3x}=\frac{-2y(xy-2)}{3x(xy+5)}=\frac{4y-2xy^{2}}{15x+3x^{2}y} [/tex]
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Assumption 2: If your complex fraction is given as below
[tex] \frac{\frac{-2}{x} +\frac{5}{y} }{\frac{3}{y} -\frac{2}{x} } [/tex]
The LCD of both the top part and bottom part is 'xy', we can multiply that in both top and bottom part to simplify the given complex fraction as below:
[tex] \frac{xy(\frac{-2}{x}+\frac{5}{y}) }{xy(\frac{3}{y}-\frac{2}{x}) } =\frac{\frac{-2xy}{x}+\frac{5xy}{y}}{\frac{3xy}{y}-\frac{2xy}{x}} [/tex]
Cancelling the common factors in both top and bottom of each fraction we will get the below simplified fraction
[tex] \frac{-2y+5x}{3x-2y} =\frac{5x-2y}{3x-2y} [/tex]
The expression expressed as a single fraction is expressed as [tex]=\frac{-2y(yx-2)}{3x(yx+5)}[/tex]
Given the expression:
[tex]\dfrac{\frac{-2}{x+\frac{5}{y} } }{\frac{3}{y-\frac{2}{x} } }[/tex]
First, we need to write the denominators as a single fraction as shown:
[tex]x+\frac{5}{y} = \frac{yx+5}{y}[/tex]
Similarly for [tex]y-\frac{2}{x}=\frac{yx-2}{x}[/tex]
Substitute the resulting function into the given function, we will have:
[tex]=\dfrac{\frac{-2}{\frac{yx+5}{y} } }{\frac{3}{\frac{yx-2}{x} } }\\\\=\dfrac{\frac{-2y}{yx+5} }{\frac{3x}{yx-2} } \\=\frac{-2y}{yx+5}\times\frac{yx-2}{3x} \\=\frac{-2y(yx-2)}{3x(yx+5)}[/tex]
This shows that the expression expressed as a single fraction is expressed as [tex]=\frac{-2y(yx-2)}{3x(yx+5)}[/tex]
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