Thực hiện phép tính \(\frac{{ - 4}}{3} - \frac{{12}}{5}\).
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-5/12 . 2/7 + 7/12 . -3/4
= -10/84 + -21/48
= -5/42 + -7/16
= -40/336 + -147/336
= -187/336
\(\frac{-5}{12}\).\(\frac{2}{7}\)+\(\frac{7}{12}\).\(\frac{-3}{4}\)
=\(\frac{5}{12}\).\(\frac{-2}{7}\)+\(\frac{7}{12}\).\(\frac{-3}{4}\)
=\(\frac{5}{12}\).(\(\frac{-2}{7}\)+\(\frac{-3}{4}\))
\(\frac{5}{12}\).\(\frac{-29}{28}\)
=\(\frac{-145}{336}\)
((nho k cho mink nhe)
1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60.
Thừa số phụ:
60:12 =5; 60:15=4
Ta được:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)
\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)
b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252.
Thừa số phụ:
252:7 = 36; 252:9 = 28; 252:12 = 21
Ta được:
\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)
\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)
\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)
2. a) Ta có BCNN(8, 24) = 24 nên:
\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)
b) Ta có BCNN(12, 16) = 48 nên:
\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).
\(6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(=\frac{77}{12}:\frac{11}{4}+\frac{45}{4}.\left(\frac{5}{15}-\frac{3}{15}\right)\)
\(=\frac{77}{12}.\frac{4}{11}+\frac{45}{4}.\frac{2}{15}\)
\(=\frac{7}{3}+\frac{3}{2}\)
\(=\frac{14}{6}+\frac{9}{6}\)
\(=\frac{23}{6}=3\frac{5}{6}\)
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
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)
Bài này dễ mà bn
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
Bài này về tỷ số
Khó lắm bạn ơi
Thông cảm cho mình
~~~ Chúc bạn học giỏi ~~~
\(\begin{array}{l}\frac{{ - 4}}{3} - \frac{{12}}{5} = \frac{{ - 4}}{3} + \frac{{ - 12}}{5}\\ = \frac{{ - 20}}{{15}} + \frac{{ - 36}}{{15}} = \frac{{ - 56}}{{15}}\end{array}\).