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:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

\(\left(-\frac{5}{12}\right):\frac{7}{3}-\left(-\frac{5}{12}\right):\frac{7}{4}=\left(-\frac{5}{12}\right):\left(\frac{7}{3}-\frac{7}{4}\right)=\left(-\frac{5}{12}\right):\frac{7}{12}=-\frac{5}{7}\)

\(\left[\left(\frac{2}{5}\right)^0\right].\frac{19}{13}-\left(\frac{7}{3}\right)^{2019}.\frac{3}{7}^{2019}\)

\(=\left(\frac{2}{5}\right)^0.\frac{19}{13}-\left(\frac{7}{3}.\frac{3}{7}\right)^{2019}\)

\(=1.\frac{19}{13}-1^{2019}\)

\(=1.\frac{19}{13}-1\)

\(=\frac{19}{13}-1\)

\(=\frac{6}{13}\)

                                                            Bài giải

a, \(\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)

\(=\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)

\(=\left(-\frac{5}{12}\right)\cdot\frac{3}{7}-\left(-\frac{5}{12}\right)\cdot\frac{4}{7}\)

\(=\frac{-15}{84}+\frac{20}{84}=\frac{5}{84}\)

b, \(\left[\left(\frac{2}{5}\right)^0\right]^{2020}\cdot\frac{19}{37}-\left(\frac{7}{3}\right)^{2019}\cdot\frac{3^{2019}}{7}\)

\(=1^{2020}\cdot\frac{19}{37}-\frac{7^{2019}}{3^{2019}}\cdot\frac{3^{2019}}{7}\)

\(=\frac{19}{37}-7^{2018}\)

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

mình tính được -60 ?

a) \(\left(\frac{-1}{6}+\frac{5}{-12}\right)+\frac{7}{12}=\left(\frac{-2}{12}+\frac{-5}{12}\right)+\frac{7}{12}=\left(\frac{-7}{12}\right)+\frac{7}{12}=0\)

b)\(\frac{7}{36}-\frac{8}{-9}+\frac{-2}{3}=\frac{7}{36}+\frac{32}{36}-\frac{24}{36}=\frac{15}{36}=\frac{5}{12}\)

c) \(\frac{3}{5}-\frac{2}{5}.\frac{10}{12}=\frac{3}{5}-\frac{2}{5}.\frac{5}{6}=\frac{3}{5}-\frac{1}{3}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)

d) \(\frac{2}{\left(-3\right)^2}+\frac{5}{-13}-\frac{-3}{4}=\frac{2}{9}-\frac{5}{13}+\frac{3}{4}=\frac{8}{36}-\frac{15}{36}+\frac{27}{36}=\frac{5}{9}\)

\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)

\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)

\(A=\frac{2\left(10-1\right)}{5-14}\)

\(A=\frac{2.9}{-9}\)

\(A=-2\)

Vậy \(A=-2\)

\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)

\(B=81.\frac{18}{5}.\frac{2}{9}\)

\(B=\frac{324}{5}\)

Vậy \(B=\frac{324}{5}\)

Chúc bạn học tốt ~ ( mỏi tay qué >_< ) 

\(B=81.\left[\frac{3\left(12-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}\right)}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{79.2.1001001}{79.9.1001001}\)

\(B=81.\left[3.\frac{6}{5}\right].\frac{2}{9}\)

\(B=\frac{9.9.3.6.2}{5.9}\)

\(B=\frac{9.3.6.2}{5}\)

\(B=\frac{324}{5}\)

Tick cho minh nha Quang Hai Duong tick minh may man ca nam

Kakashi_kun linh tinh vừa thôi

Bài thế kia mà bảo là toán lớp 6

a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)

\(=\frac{8+9+10}{12}\)

\(=\frac{27}{12}=\frac{9}{4}\)

b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)

\(=\frac{45-14+20}{24}\)

\(=\frac{51}{24}=\frac{17}{8}\)

2)

a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)

\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)

\(=1+\frac{7}{13}+\frac{1}{7}\)

\(=\frac{20}{13}+\frac{1}{7}\)

\(=\frac{153}{91}\)

Tí tớ trả lời tiếp