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Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
Tính nhanh
19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 25 + 26
1/3 + 1/4 + 1/5 + 4/6 + 9/12 + 16/20
\(19+18+17+16+14+21+22+23+24+25+26\)
\(=\left(19+21\right)+\left(18+22\right)+\left(17+23\right)+\left(16+24\right)+\left(14+26\right)+25\)
\(=30+30+30+30+30+25\)
\(=175\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{6}+\dfrac{9}{12}+\dfrac{16}{20}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)
\(\text{=}1+1+1\)
\(\text{=}3\)
Ta có:\(x.\frac{1}{5}+x.\frac{4}{5}=2\)
=>\(x.\frac{1}{5}+\frac{4}{5}=2\)
=>\(x.\frac{5}{5}=2\)
=>\(x.1=2\)
=>\(x=2:1\)
=>\(x=2\)
Ta có : 21.5 - 21 - 4.21 = 21.5 - 21.1 - 21.4 = 21(5 - 1 - 4) = 21.0 = 0
=> \(\frac{1+2+3+4+5+6+7}{21\cdot5-21-4\cdot21}=\frac{1+2+3+4+5+6+7}{0}\)
=> không thể tính được vì mẫu số phải khác 0
câu A cách 1
\(\dfrac{17}{19}\times\dfrac{12}{15}\times\dfrac{19}{17}\)
=\(\dfrac{17\times12\times19}{19\times15\times17}\)
=\(\dfrac{12}{15}\)
câu A cách 2
\(\dfrac{17}{19}\times\dfrac{12}{15}\times\dfrac{19}{17}\)
=\(\dfrac{17}{19}\times\dfrac{19}{17}\times\dfrac{12}{15}\)
=1 x \(\dfrac{12}{15}\)
=\(\dfrac{12}{15}\)
a) \(\dfrac{7}{9}\times\dfrac{13}{25}\times\dfrac{9}{7}=\left(\dfrac{7}{9}\times\dfrac{9}{7}\right)\times\dfrac{13}{25}=1\times\dfrac{13}{25}=\dfrac{13}{25}\)
b) \(\dfrac{5}{9}\times\dfrac{21}{25}+\dfrac{21}{25}\times\dfrac{4}{9}\)
\(=\dfrac{21}{25}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\)
\(=\dfrac{21}{25}\times1=\dfrac{21}{25}\)
c) \(\dfrac{19}{5}\times\dfrac{11}{16}-\dfrac{9}{5}\times\dfrac{11}{16}\)
\(=\dfrac{11}{16}\times\left(\dfrac{19}{5}-\dfrac{9}{5}\right)\)
\(=\dfrac{11}{16}\times2=\dfrac{22}{16}=\dfrac{11}{8}\)
bằng 14 nhé.................................
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