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a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
Mik làm Bài 2 nhé ~
Bài 2 :
a) \(x-\dfrac{1}{2}=-\dfrac{1}{10}\)
\(x=-\dfrac{1}{10}+\dfrac{1}{2}\)
\(x=\dfrac{2}{5}\)
b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)
\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)
\(\dfrac{2}{3}x=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}:\dfrac{2}{3}\)
\(x=\dfrac{11}{3}.\dfrac{3}{2}\)
\(x=\dfrac{11}{2}\)
c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=2,5-\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{5}{2}-\dfrac{3}{4}\)
\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\dfrac{1}{8}x=\dfrac{5}{4}\)
\(x=10\)
Bài 1:
a) \(\dfrac{-4}{11}.\dfrac{7}{9}+\dfrac{-4}{11}.\dfrac{2}{9}-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.1-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}-\dfrac{7}{11}\)
\(=-1\)
b) \(\dfrac{3}{5}:\dfrac{-7}{10}+0,5-\left(\dfrac{-9}{14}\right)\)
\(=\dfrac{-6}{7}+\dfrac{1}{2}+\dfrac{9}{14}\)
\(=\dfrac{2}{7}\)
c) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:6\)
\(=\dfrac{3}{5}-\dfrac{4}{15}\)
\(=\dfrac{1}{3}\)
a) \(\dfrac{-4}{3}-\dfrac{2}{7}+\dfrac{4}{3}\)
\(=\left(\dfrac{-4}{3}+\dfrac{4}{3}\right)-\dfrac{2}{7}\)
\(=0-\dfrac{2}{7}=\dfrac{-2}{7}\)
b) \(\dfrac{1}{2}+\dfrac{-4}{5}-\dfrac{3}{10}\)
\(=\dfrac{5}{10}+\dfrac{-8}{10}-\dfrac{3}{10}\)
\(=\dfrac{5+\left(-8\right)-3}{10}=\dfrac{-3}{5}\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
c) \(\dfrac{11}{10}-\dfrac{-7}{24}=\dfrac{11}{10}+\dfrac{7}{24}=\dfrac{167}{120}\)
e) \(\dfrac{-8}{3}\cdot\dfrac{15}{7}=\dfrac{-120}{21}=\dfrac{-40}{7}\)
f) \(\dfrac{-2}{5}\cdot4\dfrac{1}{2}=\dfrac{-2}{5}\cdot\dfrac{9}{2}=-\dfrac{9}{5}\)
g) \(\dfrac{5}{3}:\dfrac{5}{-3}=\dfrac{5}{3}:\dfrac{-5}{3}=\dfrac{5}{3}\cdot\dfrac{-3}{5}=-1\)
h) \(\dfrac{5}{4}:\left(-9\right)=\dfrac{5}{4}:\dfrac{-9}{1}=\dfrac{5}{4}\cdot\dfrac{-1}{9}=-\dfrac{5}{36}\)
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
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, \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{4}{13}.\dfrac{6}{7}\)
= \(\dfrac{3}{4}+\dfrac{6}{5}-1\) = \(\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)\)
= \(\dfrac{15}{20}+\dfrac{24}{20}-\dfrac{20}{20}\) = \(\dfrac{6}{7}.\left(\dfrac{17}{13}-\dfrac{4}{13}\right)\)
= \(\dfrac{39}{20}-\dfrac{20}{20}\) = \(\dfrac{6}{7}.1\)
= \(\dfrac{19}{20}\) = \(\dfrac{6}{7}\)
a: \(\left(-25,5\right):5=\dfrac{-25,5}{5}=-5,1\)
b: \(1\dfrac{3}{2}-0,25+\dfrac{10}{9}\cdot\dfrac{3}{5}\)
\(=\dfrac{5}{2}-0,25+\dfrac{30}{45}\)
\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{9}{4}+\dfrac{2}{3}=\dfrac{27}{12}+\dfrac{8}{12}=\dfrac{35}{12}\)
\(a,\left(-25,5\right):5\)
\(=\left(-5,1\right)\)
\(b,1\dfrac{3}{2}-0,25+\dfrac{10}{9}.\dfrac{3}{5}\)
\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{10}{9}.\dfrac{3}{5}\)
\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}\)
\(=\dfrac{10}{4}-\dfrac{1}{4}+\dfrac{2}{3}\)
\(=\dfrac{9}{4}+\dfrac{2}{3}\)
\(=\dfrac{35}{12}\)