Verified Answer :

naǫ

\(\frac{u}{x}+\frac{5u}{x}-\frac{u}{5x}=\frac{5u}{5x}+\frac{25u}{5x}-\frac{u}{5x}=\frac{5u+25u-u}{5x}=\frac{29u}{5x}\)

\(\frac{u}{x}+\frac{5u}{x}-\frac{u}{5x}=\frac{u+5u}{x}-\frac{u}{5x}=\frac{6u}{x}-\frac{u}{5x}=\frac{6u\times5}{x\times5}-\frac{u}{5x}\\\\=\frac{30u}{5x}-\frac{u}{5x}=\frac{30u-u}{5x}=\boxed{\frac{29u}{5x}}\)

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