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a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
B = \(\frac{1}{2.2}+\frac{1}{3.3}+....+\frac{1}{9.9}\)
ta có B > \(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\) ( tự giải thích )
=> B > \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
=> B > \(\frac{2}{5}\) (1)
Ta có B < \(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}\)
=> B < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
=> B < \(\frac{8}{9}\) (2)
Từ (1) và (2) => \(\frac{2}{5}< B< \frac{8}{9}\)
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
\(\frac{12}{5}-\frac{5}{2}-\frac{1}{4}< x-1< \frac{9}{2}-\frac{3}{4}+\frac{5}{8}\)
\(\Leftrightarrow\frac{48}{20}-\frac{50}{20}-\frac{5}{20}< x-1< \frac{36}{8}-\frac{6}{8}+\frac{5}{8}\)
\(\Leftrightarrow-\frac{7}{20}< x-1< \frac{35}{8}\)
\(\Leftrightarrow-0,35< x-1< 4,375\)
Tìm x nốt nhé
Ta có : S =\(\frac{1}{2^2}\)\(+\)\(\frac{1}{3^2}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
= \(\frac{1}{2.2}\)\(+\)\(\frac{1}{3.3}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
\(\Rightarrow\)S > \(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)...\(+\)\(\frac{1}{9.10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)..\(+\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{10}\)
\(\Rightarrow\)S < \(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)...\(+\)\(\frac{1}{8.9}\)
=\(1\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)...\(+\)\(\frac{1}{8}\)\(-\)\(\frac{1}{9}\)
= \(1\)\(-\)\(\frac{1}{9}\)= \(\frac{8}{9}\)
Vậy \(\frac{2}{5}\)< S < \(\frac{8}{9}\)(đpcm)
Chúc bạn học tốt
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}\)
\(S>\frac{4}{10}=\frac{2}{5}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)
=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)