27/23+5/15+3/7-4/23-12/9
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VH
1 tháng 4 2023
Bài 2. > , < , = ?
5/7 < 4/3 2/5 < 6/10 1/4 = 3/12 27/36 >
2/9 7/6 > 7/9
7/2 = 2/7 15/23 < 1 27/9 > 2 14/15 < 1 51/17 < 4
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
NK
4
10 tháng 3 2020
\(\frac{21}{8}>\frac{5}{8}\)
\(\frac{3}{4}=\frac{12}{16}\)
\(\frac{29}{15}>\frac{7}{15}\)
\(\frac{157}{23}>\frac{103}{23}\)
\(\frac{1}{3}=\frac{9}{27}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
BA
5
8 tháng 3 2023
\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)
\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)
\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)
\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)
\(\frac{27}{23}\)+ \(\frac{5}{15}\)+ \(\frac{3}{7}\)-\(\frac{4}{23}\)- \(\frac{12}{9}\)= \(\frac{27}{23}\)+ \(\frac{1}{3}\)+ \(\frac{3}{7}\)-\(\frac{4}{23}\)- \(\frac{4}{3}\)
= (\(\frac{27}{23}\)- \(\frac{4}{23}\)) + (\(\frac{1}{3}\)- \(\frac{4}{3}\)) + \(\frac{3}{7}\)
= 1+ (-1) + \(\frac{3}{7}\)= \(\frac{3}{7}\)