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Ta có :
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...........+\dfrac{1}{19}\)
\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+.......+\dfrac{1}{19}\right)\)
Ta thấy :
\(\dfrac{1}{5}>\dfrac{1}{20}\)
\(\dfrac{1}{6}>\dfrac{1}{20}\)
..................
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\Rightarrow B>\dfrac{1}{4}+\left(\dfrac{1}{20}+\dfrac{1}{20}+.........+\dfrac{1}{20}\right)\)(\(15\) p/s \(\dfrac{1}{20}\))
\(B>\dfrac{1}{4}+\dfrac{1}{20}.15\)
\(B>\dfrac{1}{4}+\dfrac{3}{4}=1\Rightarrow B>1\rightarrowđpcm\)
~ Học tốt ~
\(B=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}\\ B=1+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\\ B>1+\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ B>1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ B>4\)
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)
\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)
\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{10}\)
\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)
\(\Leftrightarrow D< 1\left(đpcm\right)\)
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
Có:
\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)
\(A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)
Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
...
\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)
\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{65}{132}\)
Chúc bạn học tốt!
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\right)\)
Các phân số \(\dfrac{1}{5}\), \(\dfrac{1}{6}\), \(\dfrac{1}{7}\), ..., \(\dfrac{1}{19}\) đều lớn hơn \(\dfrac{1}{20}\), tất cả có 15 phân số nên:
\(B>\dfrac{1}{4}+\left(\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Vậy B > 1
e! Chung minh di tai sao lai lam the : phai co ly do chu( ko phai cu thich la ko lam ngay duoc dau