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a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
\(=1.\frac{7}{12}:\frac{7}{12}\)
\(=1\)
b.
\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Tk mk nha!
b) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
a) Ta có: BCNN(15,10) = 30 nên ta chọn mẫu số chung là 30
\(\frac{11}{15}+\frac{9}{10}=\frac{22}{30}+\frac{27}{30}=\frac{49}{30}\)
b) Ta có: BCNN(6,9,12) = 36 nên ta chọn mẫu số chung là 36
\(\frac{5}{6} + \frac{7}{9} + \frac{{11}}{{12}} = \frac{{30}}{{36}} + \frac{{28}}{{36}} + \frac{{33}}{{36}} = \frac{{91}}{{36}}\)
c) Ta có: BCNN(24,21) = 168 nên ta chọn mẫu số chung là 168
\(\frac{7}{{24}} - \frac{2}{{21}} = \frac{{49}}{{168}} - \frac{{16}}{{168}} = \frac{{33}}{{168}}=\frac{11}{56}\)
d) Ta có: BCNN(36,24) = 72 nên ta chọn mẫu số chung là 72
\(\frac{{11}}{{36}} - \frac{7}{{24}} = \frac{{22}}{{72}} - \frac{{21}}{{72}} = \frac{1}{{72}}\)
1-2+3-4+5-6+...+2011-2012
=2012-2011+...+6-5+4-3+2-1
=(2012-2001)+...+(6-5)+(4-3)+(2-1)
=1+1+1+...+1+1(có 1006 số 1)
=1x60
=60
\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)
\(A=1+\frac{1}{2}\)
\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)
\(A=\frac{3}{2}\)
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
a ) \(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{-12}{19}\)
\(=\frac{7}{19}.\frac{11}{11}+\frac{-12}{19}\)
\(=\frac{7}{19}.1+\frac{-12}{19}\)
\(=\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7+\left(-12\right)}{19}\)
\(=-\frac{5}{19}\)
b ) \(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{-7.39.50}{25.-14.78}=\frac{-1.1.2}{1.-2.2}=\frac{-2}{-4}=\frac{1}{2}\)
c ) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{3}{4}\)
a)\(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}=\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{7}{19}.\frac{-12}{7}=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}+-\frac{12}{7}\right)=\frac{7}{19}.\left(\frac{-5}{7}\right)=-\frac{5}{19}\)
b)\(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{\left(-7\right).39.50}{25.\left(-14\right).78}=\frac{\left(-7\right).3.13.2.5.5}{5.5.\left(-7\right).2.2.13.3}=\frac{1}{2}\)
c)\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{2}{7}+\frac{1}{2}=\frac{11}{14}\)
\(6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(=\frac{77}{12}:\frac{11}{4}+\frac{45}{4}.\left(\frac{5}{15}-\frac{3}{15}\right)\)
\(=\frac{77}{12}.\frac{4}{11}+\frac{45}{4}.\frac{2}{15}\)
\(=\frac{7}{3}+\frac{3}{2}\)
\(=\frac{14}{6}+\frac{9}{6}\)
\(=\frac{23}{6}=3\frac{5}{6}\)