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Lời giải:
1.
$3\frac{1}{4}-2\frac{1}{3}=3+\frac{1}{4}-(2+\frac{1}{3})$
$=3-2+\frac{1}{4}-\frac{1}{3}$
$=1+\frac{1}{4}-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}$
2.
$6\frac{5}{9}+2\frac{5}{6}=6+\frac{5}{9}+2+\frac{5}{6}=8+\frac{5}{6}+\frac{5}{9}=8+\frac{25}{18}=8+1+\frac{7}{18}=9+\frac{7}{18}=9\frac{7}{18}$
3.
$6\frac{5}{9}-2\frac{5}{6}=6+\frac{5}{9}-(2+\frac{5}{6})$
$=6+\frac{5}{9}-2-\frac{5}{6}$
$=(6-2)+\frac{5}{9}-\frac{5}{6}$
$=4+\frac{5}{9}-\frac{5}{6}=\frac{41}{9}-\frac{5}{6}=\frac{67}{8}$
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
Bài 1:
\(\left(\frac{9}{4}:\frac{6}{5}\right)-1=\frac{45}{24}-1=\frac{15}{8}-1=\frac{7}{8}\)
\(2+\left(\frac{1}{4}\times\frac{2}{5}\right)=2+\frac{2}{10}=2+\frac{1}{5}=\frac{11}{5}\)
Bài 2:
\(\frac{3}{5}+\frac{2}{3}=\frac{9}{15}+\frac{10}{15}=\frac{19}{15}\)
\(1-\frac{9}{11}=\frac{1}{1}-\frac{9}{11}=\frac{11}{11}-\frac{9}{11}=\frac{2}{11}\)
\(\frac{7}{9}\times\frac{3}{14}=\frac{21}{126}=\frac{3}{18}\)
\(\frac{15}{7}:\frac{5}{21}=\frac{315}{35}=9\)
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
\(\frac{5}{8}+\frac{2}{9}-\frac{2}{5}+\frac{3}{8}+\frac{4}{9}+\frac{1}{3}-\frac{3}{5}\)
\(=\frac{5}{8}+\frac{2}{9}-\frac{2}{5}+\frac{3}{8}+\frac{4}{9}+\frac{3}{9}-\frac{3}{5}\)
\(=\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{2}{9}+\frac{4}{9}+\frac{3}{9}\right)-\left(\frac{2}{5}+\frac{3}{5}\right)\)
\(=\frac{8}{8}+\frac{9}{9}-\frac{5}{5}\)
\(=1+1-1\)
\(=2-1\)
\(=1\)
1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
(5/2-1/3)*9/2
=13/6*9/2
=39/4
\(\left(\frac{5}{2}-\frac{1}{3}\right)\times\frac{9}{2}\)
\(=\frac{13}{6}\times\frac{9}{2}\)
\(=\frac{39}{4}.\)