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\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{870}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}\)
\(=\frac{29}{30}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
bn ơi ko thì bn làm từng phép một cũng đc nhé các bn giải hộ mình mai mình phải nộp rồi
a=113/13-(24/7+53/13)=113/13-24/7-53/13=60/13-24/7=396/16
b=(64/9+37/11)-44/9=64/9+37/11-44/9=20/9+37/11=181/33
c=-5/7(2/11+9/11)+15/7=-5/7+15/7=10/7
d=0.7.22/3.20.0.375.5/28=0
e=(6,17+35/9-236/97)(1/3-0.25-1/12)=A.(1/3-1/4-1/12)=A.(1/12-1/12)=A.0=0
e=(6,17 + 3 5/9 - 2 36/97 ) . ( 1/3 - 0,25 - 1/12)=(6,17 + 3 5/9 - 2 36/97 ) .(1/3-1/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(4/12-1/12-0,25)=(6,17 + 3 5/9 - 2 36/97 ) .(3/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(1/4-0.25)=(6,17 + 3 5/9 - 2 36/97 ) .(0.25-0.25)
=(6,17 + 3 5/9 - 2 36/97 ) .0=0
Vậy E=0
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
1)\(1\frac{3}{4}-1\frac{3}{8}:\left(\frac{3}{8}+\frac{9}{20}\right)=\frac{7}{4}-\frac{11}{8}:\frac{33}{40}=\frac{7}{4}-\frac{5}{3}=\frac{1}{12}\)
\(1,=\dfrac{1}{12}-\dfrac{4}{12}=\dfrac{3}{12}\\ 2,=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\\ 3,\dfrac{-1}{6}-\dfrac{2}{5}=\dfrac{-5}{30}-\dfrac{12}{30}=\dfrac{-17}{30}\\ 4,=\dfrac{15}{20}-\dfrac{1}{20}=\dfrac{14}{20}\)
1,=112−412=3122,=636−136=5363,−16−25=−530−1230=−17304,=1520−120=1420