\(\frac{x}{2^2}+\frac{x}{3^2}+\frac{x}{4^2}=\frac{x}{2^3}+\frac{x}{3^3}+\frac{x}{4^3}\)

\(\Rightarrow x.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\right)=x.\left(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}\right)\)

Mà \(\frac{1}{2^2}>\frac{1}{2^3};\frac{1}{3^2}>\frac{1}{3^3};\frac{1}{4^2}>\frac{1}{4^3}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\ne\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}\)

=> x = 0

Vậy x = 0

3 số chắc chắn 100% mình làm rồi