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\(S=\dfrac{1}{30}+\dfrac{2}{48}+\dfrac{3}{88}+\dfrac{4}{165}+\dfrac{5}{300}\\ S=\dfrac{1}{5.6}+\dfrac{2}{6.8}+\dfrac{3}{8.11}+\dfrac{4}{11.15}+\dfrac{5}{15.20}\\ S=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}\\ S=\dfrac{1}{5}-\dfrac{1}{20}\\ S=\dfrac{4}{20}-\dfrac{1}{20}\\ S=\dfrac{3}{20}\)
\(a,=\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}=\dfrac{45}{10}=4,5\\ b,=\dfrac{4}{5}\times\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\times2=\dfrac{8}{5}\times\dfrac{1}{8}=\dfrac{1}{5}=0,2\)
a) Rút gọn các phân số về tối giản, ta được:
\(\dfrac{1}{10}\)+\(\dfrac{2}{10}\)+\(\dfrac{3}{10}\)+\(\dfrac{4}{10}\)+\(\dfrac{5}{10}\)+\(\dfrac{6}{10}\)+\(\dfrac{7}{10}\)+\(\dfrac{8}{10}\)+\(\dfrac{9}{10}\)= kết quả là \(\dfrac{45}{10}\) ra số thập phân = \(4,5\)
b) \(\dfrac{4}{5}\) \(\times\) \(\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\) = \(\dfrac{4}{5}\times\dfrac{1}{8}\) = \(\dfrac{4}{40}=\dfrac{1}{10}\)\(\div\dfrac{1}{2}\)
= \(\dfrac{2}{10}=0,2\)
a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)
\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)
49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)
b) Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)
\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)
\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))
\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)
Lại có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
Mà :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)
\(\Rightarrow A>48\) \(\rightarrowđpcm\)
a) S
b) Đ
c) Đ
d) S
A sai
B đúng
C sai