Đặt: \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\)

Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\) 

\(\Leftrightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A< \frac{99}{100}< 1\)

5dm 43cm = \(\frac{5}{10}\)m \(+\)\(\frac{43}{100}\)m \(=\)\(\frac{50+43}{100}\)m\(=\)\(\frac{93}{100}\)m \(=0,93\)m

Vậy :

5dm 43cm = 0,93m .

46489cm khoi = 46,489 dm khoi = 0,046489m khoi

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