How do you evaluate #10\div (2+\sqrt{2})#?

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Answer 1 · 2017-09-17T07:57:35

#=2-sqrt2#

#10/(2+sqrt2)#

#=10/(2+sqrt2)*(2-sqrt2)/(2-sqrt2)#

#=(10(2-sqrt2))/(2+sqrt2(2-sqrt2))#

#=(20-10sqrt2)/(4-2)#

#=(20-10sqrt2)/2#

#=10-5sqrt2#

#=2-sqrt2#

Hope this helps!