Question

Biết a^2 +b^2 =1 , a^4 + b^4 =1/2 .tính a^2020 + b^2020

Answer 1

(a²+b²)²=1
<=> a⁴+b⁴+2a²b²=1
<=> 2a²b²=1/2
<=> ab=1/2

Có a²+b²-2ab=1-1 <=> (a-b)^2=0 <=> a=b
Mặt khác a²+b²+2ab=2 <=> (a+b)^2 =2 <=> 4a^2=2 <=>a= \(\dfrac{\sqrt{2}}{2}\)

Có a²⁰²⁰+b²⁰²⁰= 2a²⁰²⁰= 2(\(\dfrac{\sqrt{2}}{2}\))^2.1010=2(\(\dfrac{1}{2}\))^{1010=\(\dfrac{2.1}{2.2^{2009}}\)=\(\dfrac{1}{2^{2009}}\)}