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Ta có: \(1+3+6+10+\cdots+45+55\)
\(=\frac12\times\left(2+6+12+20+\cdots+90+110\right)\)
\(=\frac12\times\left(1\times2+2\times3+\cdots+9\times10+10\times11\right)\)
\(=\frac12\times\left\lbrack1\times\left(1+1\right)+2\times\left(2+1\right)+\cdots+10\times\left(10+1\right)\right\rbrack\)
\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+10\times10\right)+\left(1+2+\cdots+10\right)\right\rbrack\)
\(=\frac12\times\left\lbrack10\times\left(10+1\right)\times\frac{\left(2\times10+1\right)}{6}+10\times\frac{11}{2}\right\rbrack\)
\(=\frac12\times\left\lbrack10\times11\times\frac{21}{6}+5\times11\right\rbrack=\frac12\times\left\lbrack5\times11\times7+5\times11\right\rbrack\)
\(=\frac12\times5\times11\times\left(7+1\right)=\frac{55}{2}\times8=55\times4=220\)
Ta có: \(1\times10+9\times2+\cdots+10\times1\)
\(=2\times\left(1\times10+2\times9+3\times8+4\times7+5\times6\right)\)
\(=2\times\left\lbrack1\times\left(11-1\right)+2\times\left(11-2\right)+3\times\left(11-3\right)+4\times\left(11-4\right)+5\times\left(11-5\right)\right\rbrack\)
\(=2\times\left\lbrack11\times\left(1+2+3+4+5\right)-\left(1\times1+2\times2+3\times3+4\times4+5\times5\right)\right\rbrack\)
\(=2\times\left\lbrack11\times15-\left(1+4+9+16+25\right)\right\rbrack\)
=2x(165-55)
=2x110
=220
Ta có: \(\frac{1\times10+9\times2+\cdots+10\times1}{1+3+6+10+\cdots+45+55}\)
\(=\frac{220}{220}\)
=1
\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{3+2}{6}-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{1}{5}=\dfrac{5\times5-6}{30}=\dfrac{19}{30}\\ \dfrac{1}{5}:4+\dfrac{3}{4}=\dfrac{1}{5}\times4+\dfrac{3}{4}=\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{4\times4+3\times5}{20}=\dfrac{31}{20}\)
\(\dfrac{4}{3}\times\dfrac{9}{5}-\dfrac{3}{10}=\dfrac{36}{15}-\dfrac{3}{10}=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{12\times2-3}{10}=\dfrac{21}{10}\)
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)
= \(\dfrac{1\times15}{2\times15}\) + \(\dfrac{1\times10}{3\times10}\) - \(\dfrac{1\times6}{5\times6}\)
= \(\dfrac{15}{30}\) + \(\dfrac{10}{30}\) - \(\dfrac{6}{30}\)
= \(\dfrac{19}{30}\)
\(\dfrac{1}{5}\) : 4 + \(\dfrac{3}{4}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
= \(\dfrac{1}{20}\) + \(\dfrac{3\times5}{4\times5}\)
= \(\dfrac{1}{20}+\dfrac{15}{20}\)
= \(\dfrac{16}{20}\)
= \(\dfrac{4}{5}\)
\(\dfrac{4}{3}\) \(\times\) \(\dfrac{9}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{12}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{24}{10}\) - \(\dfrac{3}{10}\)
= \(\dfrac{21}{10}\)
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
a) \(\left(\dfrac{1}{4}+\dfrac{1}{12}\right):\dfrac{1}{13}\)
\(=\left(\dfrac{3}{12}+\dfrac{1}{12}\right):\dfrac{1}{13}\)
\(=\dfrac{4}{12}\times13\)
\(=\dfrac{1}{3}\times13\)
\(=\dfrac{13}{3}\)
b) \(\dfrac{3}{5}:\dfrac{2}{9}-\dfrac{1}{10}\)
\(=\dfrac{3}{5}\times\dfrac{9}{2}-\dfrac{1}{10}\)
\(=\dfrac{27}{10}-\dfrac{1}{10}\)
\(=\dfrac{26}{10}\)
\(=\dfrac{13}{5}\)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
a)
`2/3+5/2-3/4`
`=10/4-3/4+2/3`
`=7/4+2/3`
`=21/12+8/12`
`=29/12`
b)
`2/5xx1/2:1/3`
`=2/10xx3/1`
`=6/10=3/5`
c)
`2/9:2/9xx1/3`
`=2/9xx9/2xx1/3`
`=1xx1/3`
`=1/3`
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)
= \(\dfrac{38-9}{12}\)
= \(\dfrac{29}{12}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)
= \(\dfrac{3}{5}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
= 1 x \(\dfrac{1}{3}\)
= \(\dfrac{1}{3}\)
Ta có: \(\frac{1}{1\times10}+\frac{1}{2\times9}+\cdots+\frac{1}{10\times1}\)
\(=2\times\left(\frac{1}{1\times10}+\frac{1}{2\times9}+\cdots+\frac{1}{5\times6}\right)\)
\(=\frac{2}{11}\times\left(\frac{11}{1\times10}+\frac{11}{2\times9}+\cdots+\frac{11}{5\times6}\right)\)
\(=\frac{2}{11}\times\left(1+\frac12+\frac13+\cdots+\frac19+\frac{1}{10}\right)\)
Ta có: \(A=\frac{1+\frac12+\cdots+\frac{1}{10}}{\frac{1}{1\times10}+\frac{1}{2\times9}+\cdots+\frac{1}{9\times2}+\frac{1}{10\times1}}\)
\(=\frac{1+\frac12+\cdots+\frac{1}{10}}{\frac{2}{11}\times\left(1+\frac12+\cdots+\frac{1}{10}\right)}=1:\frac{2}{11}=\frac{11}{2}\)