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\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\frac{11}{125}-\left(\frac{17}{18}-\frac{8}{18}\right)+\left(\frac{17}{14}-\frac{10}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}+\left(\frac{1}{2}-\frac{1}{2}\right)\)= \(\frac{11}{125}\)
b) \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{5}{3}-\frac{2}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0+\frac{-1}{2}\)
= \(-2-\frac{-1}{2}=-\left(2+\frac{1}{2}\right)=-2\frac{1}{2}\)
3)
3/5 + 3/7-3/11 / 4/5 + 4/7- 4/11
= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)
= 3/4
1,
ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)
196/197 > 196/197+198
197/198 > 197/197+198
=> A>B
Ta có \(\frac{11}{4}\)> \(\frac{x}{20}\)> \(\frac{5}{11}\)
=> \(\frac{605}{220}\)> \(\frac{11x}{220}\)> \(\frac{100}{220}\)
=> 605 > 11x > 100
=> 56 > x > 9
\(\frac{11}{4}>\frac{x}{20}>\frac{5}{11}\)
\(\Leftrightarrow\frac{605}{220}>\frac{11x}{220}>\frac{100}{220}\)
\(\Leftrightarrow\frac{55}{20}>\frac{x}{20}>\frac{9,\left(09\right)}{20}\)
\(\Rightarrow x\in\left\{11;22;33;44\right\}\)
B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)
B=\(2+3\frac{7}{11}\)
B=\(5\frac{7}{11}\)
B = \(5\frac{7}{11}=\frac{62}{11}\)
C = 1
D = \(\frac{5}{2}=2\frac{1}{2}\)