Tính: 15/28*4/11+15/28*2/11+15/28*5/11
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NT
6
TQ
2
26 tháng 10 2020
\(\frac{17}{8}+\frac{28}{15}-\frac{8}{15}=\frac{17}{8}+\frac{4}{3}=\frac{83}{24}=3\frac{11}{24}\)
=> chọn a
7 tháng 8 2019
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
\(\dfrac{4}{15}-\dfrac{23}{28}-\left(-\dfrac{23}{28}+-\dfrac{11}{5}-\dfrac{29}{27}\right)-\dfrac{2}{27}\)
\(=\dfrac{4}{15}-\dfrac{23}{28}+\dfrac{23}{28}+\dfrac{11}{5}+\dfrac{29}{27}-\dfrac{2}{27}\)
\(=\left(\dfrac{4}{15}+\dfrac{11}{5}\right)+\left(-\dfrac{23}{28}+\dfrac{23}{28}\right)+\left(\dfrac{29}{27}-\dfrac{2}{27}\right)\)
\(=\left(\dfrac{4}{15}+\dfrac{33}{15}\right)+0+1\)
\(=\dfrac{37}{15}+1\)
\(=\dfrac{52}{15}\)
G
1
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
\(\dfrac{19}{28}+\dfrac{23}{47}-\dfrac{15}{8}+\dfrac{9}{28}+\dfrac{24}{47}\)
\(=\left(\dfrac{19}{28}+\dfrac{9}{28}\right)+\left(\dfrac{23}{47}+\dfrac{24}{47}\right)-\dfrac{15}{8}\)
\(=\dfrac{28}{28}+\dfrac{47}{47}-\dfrac{15}{8}\)
\(=1+1-\dfrac{15}{8}\)
\(=\dfrac{1}{8}\)
____________________
\(\dfrac{30}{11}+\dfrac{12}{35}-\dfrac{19}{11}-\dfrac{2}{35}-\dfrac{2}{7}\)
\(=\left(\dfrac{30}{11}-\dfrac{19}{11}\right)+\left(\dfrac{12}{35}-\dfrac{2}{35}\right)-\dfrac{2}{7}\)
\(=\dfrac{11}{11}+\dfrac{10}{35}-\dfrac{2}{7}\)
\(=1+\dfrac{2}{7}-\dfrac{2}{7}\)
\(=1\)
12 tháng 2 2018
a ) \(\frac{-4}{11}\)x \(\frac{5}{15}\)x \(\frac{11}{-4}\)
= \(\frac{-4.5.11}{11.15.\left(-4\right)}\)
= \(\frac{1}{3}\)