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1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
\(\dfrac{3}{4}\cdot\dfrac{7}{9}\cdot\dfrac{1}{9}\cdot\dfrac{7}{4}\)
\(=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot9\cdot9\cdot4}=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot3\cdot3\cdot9\cdot4}\)
\(=\dfrac{7\cdot1\cdot7}{4\cdot3\cdot9\cdot4}=\dfrac{49}{432}\)
\(\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{9}\cdot\dfrac{9}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}-1+\dfrac{5}{13}\right)\)
\(=\dfrac{6}{7}\cdot0\)
\(=0\)
\(2\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\dfrac{2\cdot11\cdot3\cdot9}{4\cdot121}=\dfrac{2\cdot11\cdot3\cdot9}{2\cdot2\cdot11\cdot11}\)
\(=\dfrac{3\cdot9}{2\cdot11}=\dfrac{27}{22}\)
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
e).
= \(\dfrac{9}{17}.\dfrac{3}{7}+\dfrac{9}{17}.\dfrac{4}{7}\)
= \(\dfrac{9}{17}.\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\dfrac{9}{17}.1=\dfrac{9}{17}\)
f) .
cj nghĩ đề sai đk=)? mạnh dạn đoán=)
\(B=\dfrac{\dfrac{2}{3}\left(\dfrac{17}{3}-\dfrac{4}{13}\right)+\dfrac{7}{3}}{\dfrac{9}{20}}=3:\dfrac{9}{20}=3\cdot\dfrac{20}{9}=\dfrac{60}{9}=\dfrac{20}{3}\)
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
a: \(=\dfrac{3}{4}+\dfrac{9}{5}\cdot\dfrac{2}{3}-1=\dfrac{3}{4}+\dfrac{6}{5}-1=\dfrac{19}{20}\)
b: \(=\dfrac{6}{7}\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)=\dfrac{6}{7}\cdot\dfrac{13}{13}=\dfrac{6}{7}\)
Thực hiện phép tính ( tính hợ lí nếu được)
a, \(\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\) b, \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{4}{13}.\dfrac{6}{7}\)
= \(\dfrac{3}{4}+\dfrac{6}{5}-1\) = \(\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)\)
= \(\dfrac{15}{20}+\dfrac{24}{20}-\dfrac{20}{20}\) = \(\dfrac{6}{7}.\left(\dfrac{17}{13}-\dfrac{4}{13}\right)\)
= \(\dfrac{39}{20}-\dfrac{20}{20}\) = \(\dfrac{6}{7}.1\)
= \(\dfrac{19}{20}\) = \(\dfrac{6}{7}\)
\(=\dfrac{52}{27}-\dfrac{160}{27}=\dfrac{-108}{27}=-4\)
= 52/27 - 160/27
= -108/27