\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)
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25 tháng 2 2018
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
TM
4 tháng 5 2018
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
4 tháng 5 2018
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
HQ
Hà Quang Minh
Giáo viên
2 tháng 11 2023
a) $\frac{4}{{25}}:\frac{4}{3} = \frac{4}{{25}} \times \frac{3}{4} = \frac{3}{{25}}$
b) $\frac{3}{{14}}:\frac{6}{7} = \frac{3}{{14}} \times \frac{7}{6} = \frac{{3 \times 7}}{{14 \times 6}} = \frac{{3 \times 7}}{{7 \times 2 \times 3 \times 2}} = \frac{1}{4}$
c) $\frac{{12}}{{15}}:2 = \frac{{12}}{{15}} \times \frac{1}{2} = \frac{{12 \times 1}}{{15 \times 2}} = \frac{{6 \times 2 \times 1}}{{15 \times 2}} = \frac{6}{{15}}$
d) $\frac{{21}}{8}:6 = \frac{{21}}{8} \times \frac{1}{6} = \frac{{21 \times 1}}{{8 \times 6}} = \frac{{7 \times 3 \times 1}}{{8 \times 3 \times 2}} = \frac{7}{{16}}$
7 tháng 8 2019
a . 7/12 . 6/11 + 7/12 . 5/11 - 2 7/12
= 7/12 . ( 6/11 + 5/11 ) - 31/12
= 7/12 . 1 - 31/12
= 7/12 - 31/12
= -2
b . -5/9 . -6/13 + 5/-9 . -5/13 - 5/9
= -5/9 . ( -6/13 + -5/13 ) - 5/9
= -5/9 . ( -1 ) -5/9
= 5/9 - 5/9
= 0
SO
31 tháng 8 2017
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
3 tháng 7 2018
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
22 tháng 9 2016
Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)
\(=\frac{14}{6}+\frac{1}{3}+\frac{14}{9}+\frac{7}{15}+\frac{14}{14}+\frac{6}{4}\)
\(=\frac{42+6+28}{18}+\frac{7}{15}+\frac{2}{2}+\frac{3}{2}\)
\(=\frac{76}{18}+\frac{7}{15}+\frac{5}{2}\)
\(=\frac{76}{18}+\frac{5}{2}+\frac{7}{15}\)
\(=\frac{76+45}{18}+\frac{7}{15}\)
\(=\frac{121}{18}+\frac{7}{15}\)
\(=\frac{605+42}{90}\)
\(=\frac{647}{90}\)
\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)
\(=(\frac{7}{3}+\frac{1}{3})+1+(\frac{14}{9}+\frac{7}{15})+\frac{3}{2}\)
\(=\frac{8}{3}+1+\frac{91}{45}+\frac{3}{2}\)
\(=2+\frac{2}{3}+1+2+\frac{1}{45}+1+\frac{1}{2}\)
\(=\left(\frac{2}{3}+\frac{1}{45}+\frac{1}{2}\right)+\left(2+1+2+1\right)\)
\(=1+\frac{17}{90}+6\)
\(=7+\frac{17}{90}\)
\(=7\frac{17}{90}\)