tính bằng cách hợp lí:
\(\frac{3}{5}+\frac{3}{11}+\frac{-3}{7}+\frac{-2}{97}-\frac{1}{35}-\frac{3}{4}+\frac{-23}{24}\)
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NN
17 tháng 9 2020
\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)
17 tháng 9 2020
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=-1+1+\frac{2}{97}\)
\(=\frac{2}{97}\)
HQ
Hà Quang Minh
Giáo viên
7 tháng 10 2023
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
7 tháng 10 2023
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
NN
16 tháng 7 2016
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b)\(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6\)
c)\(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)
\(=9.\left(-\frac{1}{3}\right)^3+9.\frac{1}{27}\)
\(=9.\left[\left(-\frac{1}{3}\right)^3+\frac{1}{27}\right]\)
\(=9.0=0\)
d)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
D
12 tháng 9 2017
\(=\left[\frac{3}{5}-\frac{1}{35}-\left(\frac{-3}{7}\right)\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{21}{35}-\frac{1}{35}+\frac{15}{35}\right]+\left[\frac{12}{44}-\frac{33}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{20}{35}+\frac{15}{35}\right]+\left[\frac{-21}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=1+\left(-1\right)\)
\(=0\)
NM
0
27 tháng 2 2015
a,Gọi tổng trên là A.
Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)
b,Gọi tổng trên là B
Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)
LP
29 tháng 2 2016
\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)