(5^(1/2)-2)^8. let 5^(1/2) = a & 2= b then equation become (a-b)^8=[{(a-b)^2}^2]^2= [{(5^(1/2)-2)^2}^2]^2 using (a-b)^2=a^2+b^2-2ab {(5+4-4[5^(1/2)])^2}^2 {(9-4√5)^2}^2 again (a-b)^2=a^2+b^2-2ab (81+80-72√5)^2 (161-72√5)^2 using (a-b)^2=a^2+b^2-2ab (51841-23184√5)
![ประสบการณ์ จ้าง Developer ที่ผมจะไม่มีวันลืม [ภาค2]](http://i.ytimg.com/vi/pZbsSDRrmXA/mqdefault.jpg)
(5 ➖ 2)^2^3 (2^3 ➖ 1)^1^1^1 (2^3) (x ➖ 3x+2).
(5^(1/2)-2)^8. let 5^(1/2) = a & 2= b then equation become
(a-b)^8=[{(a-b)^2}^2]^2=
[{(5^(1/2)-2)^2}^2]^2
using (a-b)^2=a^2+b^2-2ab
{(5+4-4[5^(1/2)])^2}^2
{(9-4√5)^2}^2
again (a-b)^2=a^2+b^2-2ab
(81+80-72√5)^2
(161-72√5)^2
using (a-b)^2=a^2+b^2-2ab
(51841-23184√5)
Nice Approach 👍
Minus sign should be plus in next to last step
You are right 👍
I'm really sorry 😔
That's typo.
asnwer=15 isit
asnwer=/5-2 isit
No
Esa resta no daria como resultado cero ??
No