Ta có:\(2^{30}=\left(2^3\right)^{10}=8^{10}< 9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(3^{30}=3^{20}.3^{10}< 3^{20}.4^{10}=3^{20}.\left(2^2\right)^{10}=3^{20}.2^{20}=\left(3.2\right)^{20}=6^{20}\)
\(4^{30}=4^{20}.4^{10}=4^{20}.\left(2^2\right)^{10}=4^{20}.2^{20}=\left(4.2\right)^{20}=8^{20}\)
nên \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
Ta có : x=230 + 330 + 430
x=23.10 + 33.10 + 43.10
x=(23)20 + (33)10 + (43)10
x=820 + 910 + 6410
y=320 + 620 + 820
y=32.10 + 62.10 + 82.10
y=(32)10 + (62)10 + (82)10
y=910 + 3610 + 6410
mà 910 > 820 ;3610 >910 ;6410 = 6410
nên x<y
