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b)\(4\times\dfrac{3}{8}=\dfrac{12}{8}=\dfrac{3}{2}\)
\(3:\dfrac{1}{2}\)
=\(3\times\dfrac{2}{1}\)
= 6
=\(\dfrac{1}{2}:3\)
= \(\dfrac{1}{2}\times\dfrac{1}{3}\)
= \(\dfrac{1}{6}\)
1a, \(\dfrac{3}{10}\) \(\times\) \(\dfrac{4}{9}\)
= \(\dfrac{3\times2\times2}{2\times5\times3\times3}\)
= \(\dfrac{2}{15}\)
\(\dfrac{6}{5}\): \(\dfrac{3}{7}\)
= \(\dfrac{6}{5}\) \(\times\) \(\dfrac{7}{3}\)
= \(\dfrac{14}{5}\)
\(\dfrac{3}{4}\) \(\times\) \(\dfrac{2}{5}\)
= \(\dfrac{3\times2}{2\times2\times5}\)
= \(\dfrac{3}{10}\)
\(\dfrac{5}{8}\) : \(\dfrac{1}{2}\)
= \(\dfrac{5}{8}\) \(\times\) \(\dfrac{2}{1}\)
= \(\dfrac{5}{4}\)
Bài 2 :
a, \(3+\dfrac{2}{5}\)
= \(\dfrac{15}{5}+\dfrac{2}{5}\)
= \(\dfrac{17}{5}\)
b, \(4-\dfrac{5}{7}\)
= \(\dfrac{28}{7}-\dfrac{5}{7}\)
= \(\dfrac{23}{7}\)
c, \(1-\left[\dfrac{2}{5}+\dfrac{1}{3}\right]\)
= \(1-\left[\dfrac{6}{15}+\dfrac{5}{15}\right]\)
= \(\dfrac{15}{15}-\dfrac{11}{15}\)
= \(\dfrac{4}{15}\)
Bài 2
a, 3 + \(\dfrac{2}{5}\) = \(\dfrac{15}{5}\) + \(\dfrac{2}{5}\) = \(\dfrac{17}{5}\)
b, 4 - \(\dfrac{5}{7}\) = \(\dfrac{28}{7}\) - \(\dfrac{5}{7}\) = \(\dfrac{23}{7}\)
c, 1 - ( \(\dfrac{2}{5}\) + \(\dfrac{1}{3}\))
= 1 - (\(\dfrac{6}{15}\) + \(\dfrac{5}{15}\))
= 1 - \(\dfrac{11}{15}\)
= \(\dfrac{15}{15}\) - \(\dfrac{11}{15}\)
= \(\dfrac{4}{15}\)
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
a)\(\frac{5}{6}+\frac{3}{4}x=3\Leftrightarrow\frac{3}{4}x=\frac{13}{6}\Leftrightarrow x=\frac{26}{9}\)
b)\(\frac{8}{x}\cdot\frac{3}{4}=\frac{9}{10}\Leftrightarrow\frac{8}{x}=\frac{6}{5}\Leftrightarrow x=\frac{8\cdot5}{6}=\frac{20}{3}\)
c)\(3\cdot\left(x+\frac{1}{2}\right)-\frac{1}{3}=\frac{2}{5}\Leftrightarrow3\left(x+\frac{1}{2}\right)=\frac{11}{15}\Leftrightarrow x+\frac{1}{2}=\frac{11}{45}\Leftrightarrow x=-\frac{23}{90}\)
d)\(\frac{1}{4}+\frac{x}{3}=\frac{5}{6}\Leftrightarrow\frac{x}{3}=\frac{7}{12}\Leftrightarrow x=\frac{7\cdot3}{12}=\frac{7}{4}\)
Bạn ơi , cái dấu hai chiều là sao ? trả lời nhanh cho mình nhé !
1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
làm giống như trên
c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161.(21−24501)=161.(24501225−24501)=4900153
d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
a)\(\frac{-3}{5}+\frac{1}{4}+\frac{-3}{10}\)
\(=\frac{-12}{20}+\frac{5}{20}+\frac{-6}{20}\)
\(=\frac{-13}{20}\)
b)\(\frac{1}{5}+\frac{-9}{10}+\frac{-7}{25}=\frac{10}{50}+\frac{-45}{50}+\frac{-14}{50}=\frac{-49}{50}\)
c)\(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)
=\(\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{1}{8}+\frac{-1}{8}\right)+\frac{1}{9}\)
=\(0+0+0+0+\frac{1}{9}\)
=\(\frac{1}{9}\)
b, \(\dfrac{3}{5}\) - \(\dfrac{3}{8}\)
= \(\dfrac{24}{40}\) - \(\dfrac{15}{40}\)
= \(\dfrac{9}{40}\)
c, \(\dfrac{1}{4}\) + \(\dfrac{5}{6}\)
= \(\dfrac{6}{24}\) + \(\dfrac{20}{24}\)
= \(\dfrac{26}{24}\)
= \(\dfrac{13}{12}\)