2.

a) \(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}\)

Ta có:

\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{10000}{10001}=B\)

\(\Rightarrow A.A^2< A.B\)

\(\Rightarrow A^2< \dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}=\dfrac{1}{10001}\)

\(\Rightarrow A^2< \dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(0,01\right)^2\)

\(\Rightarrow A< 0,01\)

Vậy A , 0,01

Câu 1:

a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)

\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)

\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)

\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{3}{8}\)