Evaluate #((sqrt10^1009))/(sqrt10^1011-sqrt10^1007)#without using a calculator?

2 Answers
Aug 19, 2017

#(sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007)) = 10/99 = 0.bar(10)#

Explanation:

Note that #sqrt(10) = 10^(1/2)# and if #a, b, c > 0# then:

#a^b*a^c = a^(b+c)" "# and #" "(a^b)^c = a^(bc)#

So we find:

#(sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007)) = 10^(1009/2)/(10^(1011/2)-10^(1007/2))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = 10^(1007/2+1)/(10^(1007/2+2)-10^(1007/2+0))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = color(red)(cancel(color(black)(10^(1007/2))))/color(red)(cancel(color(black)(10^(1007/2))))*(10/(100-1))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = 10/99 = 0.bar(10)#

Aug 19, 2017

Same answer, different notation.

Explanation:

#sqrt10^1009/(sqrt10^1011-sqrt10^1007)#

There is a common factor of #sqrt10^1007#.

# = (sqrt10^1007(sqrt10^2))/(sqrt10^1007(sqrt10^4-1))#

# = sqrt10^2/(sqrt10^4-1)#

# = 10/(100-1) = 10/99 = 0.bar(10)#