TH1: x=2
=>\(3^2+4^2=25=5^2\left(nhận\right)\)
TH2: x>2
\(\dfrac{3^x+4^x}{5^x}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x=1\)(1)
Vì x>2 nên \(\left(\dfrac{3}{5}\right)^x< \left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\left(\dfrac{4}{5}\right)^x< \left(\dfrac{4}{5}\right)^2=\dfrac{16}{25}\)
=>(1) vô lý
TH3: n=1
=>3+4=5(loại)
TH4: n=0
=>1+1=1(loại)