How do you simplify #\frac { \sqrt { x - 5} } { 3} - \frac { 2} { \sqrt { x - 5} }#?

Question

1254 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-15T09:02:13

#((x - 11) (sqrt (x - 5)))/(3(x - 5)#

#\frac { \sqrt { x - 5} } { 3} - \frac { 2} { \sqrt { x - 5} }#

Taking L.C.M

#rArr ((sqrt (x - 5)) (sqrt (x - 5)) - ( 2 xx 3))/((3) sqrt (x - 5))#

#rArr (sqrt ((x - 5))^2 - 6)/(3sqrt (x - 5))#

#rArr (x - 5 - 6)/(3sqrt (x - 5))#

#rArr (x - 11)/(3sqrt (x - 5))#

#rArr (x - 11)/(3sqrt (x - 5)) xx (sqrt (x - 5))/(sqrt (x - 5))#

#rArr ((x - 11) (sqrt (x - 5)))/(3sqrt (x - 5)^2#

#rArr ((x - 11) (sqrt (x - 5)))/(3 xx (x - 5)#

#rArr ((x - 11) (sqrt (x - 5)))/(3(x - 5)#

Answer 2 · 2017-08-15T10:01:32

#=(x-5 -6sqrt(x-5))/(3(x-5))#

Before we subtract the fractions, let's change the second one so there is no radical in the denominator.

#sqrt(x-5)/3 - 2/sqrt(x-5) xx sqrt(x-5)/sqrt(x-5)color(white)(xxxxx)color(blue)([ sqrt(x-5)/sqrt(x-5) =1]#

#=sqrt(x-5)/3 - (2sqrt(x-5))/((x-5))" "larr (sqrt(x-5)^2 = (x-5))#

#=(sqrt(x-5)sqrt(x-5)" " -" " 3xx2sqrt(x-5))/(3(x-5)#

#=(x-5 -6sqrt(x-5))/(3(x-5))#