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Số số hạng trong dãy số 100;98;...;2 là:
\(\dfrac{100-2}{2}+1=\dfrac{98}{2}+1=50\left(số\right)\)
Tổng của dãy số 100;98;...;2 là:
\(\left(100+2\right)\cdot\dfrac{50}{2}=102\cdot25=2550\)
100+98+...+2+97-95-93
=2550+2-93
=2552-93
=2459
a: \(\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{39}{168}-\dfrac{35}{168}+\dfrac{24}{168}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{28}{168}=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{1}{6}=\dfrac{-3}{5}+\dfrac{14}{15}\)
\(=\dfrac{-9}{15}+\dfrac{14}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
b: \(\dfrac{5}{7}\cdot\dfrac{11}{18}+\dfrac{3}{7}\cdot\dfrac{5}{18}+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\left(\dfrac{11}{7}+\dfrac{3}{7}\right)+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\cdot2+\dfrac{4}{9}=\dfrac{5}{9}+\dfrac{4}{9}=\dfrac{9}{9}=1\)
c: \(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}=10\cdot\dfrac{-7}{5}=-14\)
d: \(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
\(=\left(-\dfrac{3}{5}+\dfrac{4}{9}+\dfrac{-2}{5}+\dfrac{5}{9}\right)\cdot\dfrac{11}{7}\)
\(=\left(-\dfrac{5}{5}+\dfrac{9}{9}\right)\cdot\dfrac{11}{7}=\left(-1+1\right)\cdot\dfrac{11}{7}=0\)
e: \(\dfrac{-3}{4}\cdot5\dfrac{3}{13}-0,75\cdot\dfrac{36}{13}\)
\(=\dfrac{-3}{4}\left(5+\dfrac{3}{13}+\dfrac{36}{13}\right)\)
\(=\dfrac{-3}{4}\cdot8=-6\)
f: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{302\cdot305}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{302\cdot305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{305}\right)=\dfrac{1}{3}\cdot\dfrac{12}{61}=\dfrac{4}{61}\)
g: \(6\dfrac{5}{12}:2\dfrac{3}{4}-11\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}-\dfrac{45}{4}\cdot\dfrac{2}{15}=\dfrac{77}{12}\cdot\dfrac{4}{11}-\dfrac{3}{2}\)
\(=\dfrac{7}{3}-\dfrac{3}{2}=\dfrac{14}{6}-\dfrac{9}{6}=\dfrac{5}{6}\)
h: \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right)\cdot2\dfrac{2}{3}\cdot0,25\)
\(=\left(0,6+0,415-0,015\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)
\(=1\cdot\dfrac{2}{3}=\dfrac{2}{3}\)
i: \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right)\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{33}{20}\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{3}{2}=\dfrac{2}{2}=1\)
\(C=\left(6-\dfrac{2}{3}+\dfrac{5}{16}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{5}{16}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{5}{16}+\dfrac{7}{8}+\dfrac{5}{8}\right)-\dfrac{13}{12}\)
\(=-2-2+\left(\dfrac{5}{16}+\dfrac{12}{8}\right)-\dfrac{13}{12}\)
\(=-4-\dfrac{13}{12}+\dfrac{29}{16}=-\dfrac{157}{48}\)
\(C=\left(6-\dfrac{2}{3}-\dfrac{5}{10}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}-\dfrac{5}{10}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{7}{8}+\dfrac{5}{8}\right)+\left(\dfrac{-5}{10}-\dfrac{13}{12}\right)\)
\(=\left(-2\right)+\left(-2\right)+\dfrac{3}{2}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{18}{12}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{-1}{12}\)
\(=\dfrac{-48}{12}+\dfrac{-1}{12}\)
\(=\dfrac{-49}{12}\)
\(B=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(-\dfrac{3}{4}-\dfrac{7}{4}+\dfrac{5}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)
\(=-3+\dfrac{-5}{4}+\dfrac{-7}{5}=-3-1,25-1,4=-5,65\)
\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\\ =\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\\ =\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{5}{9}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{27}{45}+\dfrac{25}{45}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\dfrac{12}{12}+\dfrac{45}{45}+\dfrac{1}{35}\\ =1+1+\dfrac{1}{35}\\ =2+\dfrac{1}{35}\\ =\dfrac{70}{35}+\dfrac{1}{35}=\dfrac{71}{35}\)
\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{5}{9}\right)\)
\(=\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{21}{35}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{25}{45}\right)\)
\(=1+\dfrac{22}{35}+\dfrac{18}{45}\)
\(=\dfrac{315}{315}+\dfrac{198}{315}+\dfrac{126}{315}\)
\(=\dfrac{71}{35}\)
Bài 2:
a) \(A=\left\{0;1;2;3;4;5\right\}\)
\(A=\left\{x\in N|0\le x\le5\right\}\)
b) \(B=\left\{G,I,A,Đ,N,H\right\}\)
c) Tập hợp giao của A và B là C
\(\Rightarrow C=\varnothing\)
d) Tập hợp D là hợp của A và B
\(\Rightarrow D=\left\{0;1;2;3;4;5;G;I;A;Đ;N;H\right\}\)
