ta có :

17²⁰=(17²)¹⁰=289¹⁰

vì 71<289 ; 5<10 nên

71⁵<289¹⁰ hay 71⁵<17²⁰

\(\left(\sqrt{2x+3}+\sqrt{2x-3}\right)^2=\sqrt{2x-3}^2+2\sqrt{\left(2x+3\right)\left(2x-3\right)}+\sqrt{2x-3}^2=2x-3+2x+3+2\sqrt{4x^2-9}=4x+2\sqrt{4x^2-9}\)

\(\left(\sqrt{2x+y}+\sqrt{2x-y}\right)^2=2x+y+2x-y+2\sqrt{\left(2x+y\right)\left(2x-y\right)}=4x+2\sqrt{4x^2-y^2}\)