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bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
a: =1/6+14/6-3/6=12/6=2
b: =-13/8+5/4:(-5/4)
=-13/8-1=-21/8
c: =-3/8(2/5+14/5)
=-3/8*16/5
=-6/5
d: =5/34(1/4+11/9-2/9+29/4)
=5/34*(15/2+1)
=5/34*17/2
=5/4
a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)
\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)
\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)
a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)
b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)
a) \(\dfrac{3}{10}\)
b) \(\dfrac{9}{28}\)
c) \(\dfrac{88}{195}\)
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)
\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)
\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)
\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-11}{15}\)
\(\dfrac{-9}{8}\)
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
\(a,27.5^2-25.127=27.25-25.127=25\left(27-127\right)=25.\left(-100\right)=-2500\\ b,\dfrac{-5}{12}+\dfrac{3}{4}+\dfrac{1}{-3}=\dfrac{-5}{12}+\dfrac{9}{12}-\dfrac{4}{12}=\dfrac{0}{12}=0\)
a = -2500
b = 0