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a) \(\dfrac{-5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)
\(=\left(\dfrac{-5}{9}+\dfrac{2}{9}\right)-\left(\dfrac{4}{15}+\dfrac{11}{15}\right)\)
\(=\dfrac{-3}{9}-\dfrac{15}{15}\)
\(=-\dfrac{1}{3}-1\)
\(=-\dfrac{4}{3}\)
b) \(\dfrac{-8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{6}\)
\(=\left(\dfrac{-8}{13}+\dfrac{21}{13}\right)-\dfrac{7}{16}-\dfrac{1}{6}\)
\(=\dfrac{13}{13}-\dfrac{29}{48}\)
\(=1-\dfrac{29}{48}\)
\(=\dfrac{19}{48}\)
c) \(\dfrac{-15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)
\(=\dfrac{-3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)
\(=\left(\dfrac{-3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{5}{5}-\dfrac{7}{3}\)
\(=1-\dfrac{7}{3}\)
\(=-\dfrac{4}{3}\)
a: =-5/9+2/9-11/15-4/15
=-3/9-1
=-12/9=-4/3
b: =-8/13+21/13-7/16-1/6
=1-1/6-7/16
=5/6-7/16
=19/48
c: =-15/25+8/5-2/3-5/3
=-7/3+1
=-4/3
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
đáng lẻ số cuối là 127 hay 129 ko thể là số chắn [ ( 128 + 1 ) x ( 128 - 1 ) : 2 + 1 ) : 2 = ( sai đề )
A = 1 + 3 + 5 + 7 +...+ 101
Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1 = 2
Số số hạng của dãy số trên là: (101 - 1):2 +1 = 51
Tổng dãy số trên là: A = (101 + 1) x 51: 2 = 2601
(101 - 1) :2 + 1= 51
(101+1) x51:2=2601