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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
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a: \(6:\dfrac{3}{5}-1\dfrac{1}{6}\cdot\dfrac{6}{7}\)
\(=6\cdot\dfrac{5}{3}-\dfrac{7}{6}\cdot\dfrac{6}{7}\)
=10-1
=9
\(4\dfrac{1}{5}\cdot\dfrac{10}{11}+5\dfrac{2}{11}\)
\(=\dfrac{21}{5}\cdot\dfrac{10}{11}+\dfrac{57}{11}\)
\(=\dfrac{42}{11}+\dfrac{57}{11}=9\)
=>Đ
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\(a.\frac{1}{2}+\frac{7}{8}+\frac{17}{19}+\frac{42}{47}< 1+1+1+1\)
\(< 4\)
\(b.\frac{3}{2}+\frac{6}{5}+\frac{17}{11}+\frac{197}{193}>1+1+1+1\)
\(>4\)
Ủng hộ mk nha ^_-
a)Ta có:
1/2<1 7/8<1 17/19<1 42/47<1
Vậy:
1/2+7/8+17/19+42/47<1+1+1+1(do từng số hạng trong vế trái đều nhỏ hơn từng số hạng bên vế phải)
Vậy: 1/2+7/8+17/19+42/47<4
b)Ta có:
3/2>1 6/5>1 17/11>1 197/193>1
Vậy:
3/2+6/5+17/11+197/193>1+1+1+1
Vậy 3/2+6/5+17/11+197/193>4
Chúc em học tốt^^
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\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
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\(\dfrac{19}{11}+1\dfrac{3}{11}=\dfrac{19}{11}+\dfrac{14}{11}=\dfrac{33}{11}=3\)
\(2\dfrac{1}{3}:\dfrac{7}{6}=\dfrac{7}{3}\cdot\dfrac{6}{7}=\dfrac{6}{3}=2\)