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\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)
\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)
\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)
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, \(\dfrac{-5}{11}\cdot\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\cdot\dfrac{13}{17}+\dfrac{-5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\left(\dfrac{13}{17}+\dfrac{4}{17}\right)=\dfrac{-5}{11}.1=\dfrac{-5}{11}\)
Bai 1: tính nhanh A) -5/9 + 3/5 - 3/9 + -2/5 B) -5/13 + (3/5 + 3/1 - 4/10) C) 5/17 - 9/15 - 2/-17 + -2/15 D) (1/9 - 9/17) + 3/6 - ( 12/17 - 1/2) + -1/9 Bài 5: tính tổng A) 1/3 + -1/4 + 1/5 + 1/-6 + -1/-7 + 1/6 + -1/5 + 1/4 + 1/3 B) 1/12 +1/2.3+1/3.4+..+1/99100 Giúp mình nhé nhanh
c: Ta có: \(-\dfrac{5}{13}-\left(\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{4}{10}\right)\)
\(=\dfrac{-5}{13}-\dfrac{3}{5}-\dfrac{3}{13}+\dfrac{2}{5}\)
\(=\dfrac{-8}{13}-\dfrac{1}{5}\)
\(=\dfrac{-53}{65}\)
d: Ta có: \(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{5}{9}\)
\(=\dfrac{1}{9}-\dfrac{9}{17}+\dfrac{1}{2}-\dfrac{12}{17}+\dfrac{1}{2}+\dfrac{5}{9}\)
\(=\dfrac{2}{3}+1-\dfrac{21}{17}\)
\(=\dfrac{22}{51}\)
\(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\cdot\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{1-10}{6}-\dfrac{20}{3}\cdot\dfrac{8-5}{20}+\dfrac{2}{3}\cdot\dfrac{12-45}{10}\)
\(=\dfrac{-1}{2}\cdot\dfrac{-9}{6}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{66}{30}=\dfrac{-1}{4}-\dfrac{11}{5}=\dfrac{-5-44}{20}=-\dfrac{49}{20}\)
\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
b)\(B=\dfrac{3}{2}+\dfrac{13}{12}+\dfrac{31}{30}+...+\dfrac{9901}{9900}\)
\(=1+\dfrac{1}{2}+1+\dfrac{1}{12}+1+\dfrac{1}{30}+...+1+\dfrac{1}{9900}\)
\(=1+1+1+...+1\left(50cs\right)+\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(=50+\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)
\(C=\dfrac{5}{6}+\dfrac{19}{20}+\dfrac{41}{42}+...+\dfrac{10099}{10100}\)
\(=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{42}\right)+...+\left(1-\dfrac{1}{10100}\right)\)
\(=1+1+...+1\left(50cs\right)-\dfrac{1}{6}-\dfrac{1}{20}-\dfrac{1}{42}-...-\dfrac{1}{10100}\)
\(B-C=\left(50+\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)-\left(50-\dfrac{1}{6}-\dfrac{1}{20}-...-\dfrac{1}{10100}\right)\)
\(=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}+\dfrac{1}{10100}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)
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