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DochEvi [55]
3 years ago
14

arrange the expressions in the correct sequence to rationalize the denominator of the expression -(2)/(\sqrt(x+y-2)-\sqrt(x+y+2)

)
Mathematics
2 answers:
IgorC [24]3 years ago
6 0

Answer:

\frac{\sqrt{x+y-2}+\sqrt{x+y+2}}{2}

Step-by-step explanation:

Given expression :

\frac{-2}{\sqrt{x+y-2}-\sqrt{x+y+2}}

Now, we solve this expression by rationalizing method


\frac{-2}{\sqrt{x+y-2}-\sqrt{x+y+2}}\times\frac{\sqrt{x+y-2}+\sqrt{x+y+2}}{\sqrt{x+y-2}+\sqrt{x+y+2}}  


\frac{-2(\sqrt{x+y-2}+\sqrt{x+y+2})}{x+y-2-x-y-2}  

(using  (a+b)(a-b)=a^2-b^2)


\frac{-2(\sqrt{x+y-2}+\sqrt{x+y+2})}{-4}


\frac{\sqrt{x+y-2}+\sqrt{x+y+2}}{2}

this is the required arrangement which result the expression by rationalizing

ololo11 [35]3 years ago
4 0
We have to rationalize the denominator:
\frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} } = \\  \frac{-2}{ \sqrt{x+y-2} -  \sqrt{x+y+2} }* \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2}  }{ \sqrt{x+y-2}+ \sqrt{x+y+2}  }= \\  \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2})  }{x+y-2-(x+y+2)}= \\  \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2})  }{x+y-2-x-y-2}= \\  \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}  }{-4}= \\  \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2}  }{2}
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