Question

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Answer 1

Câu 5: 

\(P=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Answer 2

Câu 2.

\(x-\dfrac{3}{10}=\dfrac{17}{2}.\dfrac{1}{5}\)

\(x-\dfrac{3}{10}=\dfrac{17}{10}\)

\(x=\dfrac{17}{10}+\dfrac{3}{10}\)

\(x=\dfrac{20}{10}=2\)