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\(\dfrac{9}{8}-\dfrac{1}{2}-\dfrac{1}{6}-...........-\dfrac{1}{72}\)
\(=\dfrac{9}{8}-\left(\dfrac{1}{2}+\dfrac{1}{6}+..........+\dfrac{1}{72}\right)\)
\(=\dfrac{9}{8}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{8.9}\right)\)
\(=\dfrac{9}{8}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{9}{8}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{9}{8}-\dfrac{8}{9}\)
\(=\dfrac{17}{72}\)
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}=\dfrac{8}{9}-\left(\dfrac{1}{8.9}+\dfrac{1}{7.8}+\dfrac{1}{6.7}+\dfrac{1}{5.6}+\dfrac{1}{4.5}+\dfrac{1}{3.4}+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
a)\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}\)
=\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)=1+1=2\)
b) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
=\(\dfrac{8}{9}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
=\(\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)
thế này gọi là gian lận nha
tự đăng tự trả lời thì ko nên đăng làm gì
ko đc gì đâu 
\(C=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(\Leftrightarrow C=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(\Leftrightarrow C=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+\dfrac{1}{9.10}\right)\)
\(\Leftrightarrow C=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(\Leftrightarrow C=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(\Leftrightarrow C=\dfrac{9}{10}-\dfrac{9}{10}\)
\(\Leftrightarrow C=0\)
=>(2x-1)^2=24^2
=>2x-1=24 hoặc 2x-1=-24
=>x=-23/2 hoặc x=25/2
\(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=-\dfrac{1}{90}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=-\dfrac{1}{90}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)
\(=-\dfrac{1}{90}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=-\dfrac{1}{90}-\left(1-\dfrac{1}{9}\right)\)
\(=-\dfrac{1}{90}-\dfrac{8}{9}\)
\(=-\dfrac{9}{10}\)
1,Ta có:\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{57}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\) =\(\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+...+\dfrac{1}{2}\right)\)
= \(\dfrac{9}{10}-\left\{\dfrac{1}{\left(9.10\right)}+\dfrac{1}{\left(9.8\right)}+...+\dfrac{1}{\left(2.1\right)}\right\}\)
= \(\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{1}-\dfrac{1}{2}\right).\left(\dfrac{1}{90}=\dfrac{1}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}\right)\)=\(\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
=\(\dfrac{9}{10}-\dfrac{9}{10}\)
= 0
Ý 2 dễ rồi bạn tự tính
1, \(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{8.9}+...+\dfrac{1}{1.2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{-1}{10}+1\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
2, \(\dfrac{-5}{11}\cdot\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\cdot\dfrac{13}{17}+\dfrac{-5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\left(\dfrac{13}{17}+\dfrac{4}{17}\right)=\dfrac{-5}{11}.1=\dfrac{-5}{11}\)
\(A=\dfrac{1}{50}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{1}{50}-\dfrac{5}{14}\)
\(=\dfrac{14-250}{700}=\dfrac{-236}{700}=-\dfrac{59}{175}\)
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