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\(\frac{\left(\frac{2}{5}+1,15+\frac{-9}{20}\right):0,1}{2\frac{1}{3}+3\frac{1}{6}}=\frac{\left(\frac{2}{5}+\frac{23}{20}+\frac{-9}{20}\right):\frac{1}{10}}{\left(2+3\right)+\left(\frac{1}{3}+\frac{1}{6}\right)}=\frac{\left(\frac{2}{5}+\frac{7}{10}\right):\frac{1}{10}}{5+\frac{1}{2}}\)\(=\frac{\frac{11}{10}:\frac{1}{10}}{\frac{11}{2}}=\frac{11}{\frac{11}{2}}=11:\frac{11}{2}=2\)
a)
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)
c)
\(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)
a, \(-1,8:\left(1-\frac{3}{4}\right)\)
\(=-\frac{9}{5}:\frac{1}{4}\)
\(=-\frac{36}{5}\)
b, \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot1\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)^2\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{4}{3}-\frac{2}{7}\cdot12.25\)
\(=\frac{1}{8}\cdot\frac{4}{3}-\frac{2}{7}\cdot\frac{49}{4}\)
\(=\frac{1}{6}-\frac{7}{2}\)
\(=\frac{1}{6}-\frac{21}{6}\)
\(=-\frac{20}{6}\)
\(=-\frac{10}{3}\)
\(A=\left(27,54.\frac{11}{16}-3.405\right):3\)
\(A=\left(\frac{15147}{800}-3.405\right):3\)
\(A=\frac{12423}{800}:3\)
\(A=\frac{4141}{800}\)
\(B=\left(\frac{1}{6}+0,1+\frac{1}{5}\right):\left(\frac{1}{6}+0,1-\frac{1}{5}\right)\)
\(B=\frac{7}{15}:\frac{1}{15}\)
\(B=7\)
\(C=\left(0,5-\frac{1}{3}+0,25\right):\left(0,25-\frac{1}{6}\right)\)
\(C=\frac{5}{12}:\frac{1}{12}\)
\(C=5\)
\(D=\frac{13,5.1420+4,5.780.3}{3+6+9+...+24+27}\)
Số số hạng từ 3 đến 27 là:
(27-3):3+1=9(số hạng)
Tổng dãy số từ 3 đến 27 là:
(27+3)x9:2=135
\(D=\frac{13,5.1420+13,5.780}{135}\)
\(D=\frac{13,5.\left(1420+780\right)}{135}\)
\(D=2200\)
a) Cách 1:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \frac{2}{5} + \frac{5}{6} + \frac{4}{5}\\ = \frac{{12}}{{30}} + \frac{{25}}{{30}} + \frac{{24}}{{30}} = \frac{{61}}{{30}}\end{array}\)
Cách 2:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \left( {\frac{2}{5} + \frac{4}{5}} \right) + \frac{5}{6}\\ = \frac{6}{5} + \frac{5}{6} = \frac{{36}}{{30}} + \frac{{25}}{{30}} = \frac{{61}}{{30}}\end{array}\)
b) Cách 1:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \frac{{45}}{{60}} + \frac{{ - 44}}{{60}} + \frac{{ - 30}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\).
Cách 2:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \left( {\frac{3}{4} + \frac{{ - 1}}{2}} \right) + \frac{{ - 11}}{{15}}\\ = \left( {\frac{3}{4} + \frac{{ - 2}}{4}} \right) + \frac{{ - 11}}{{15}}\\ = \frac{1}{4} + \frac{{ - 11}}{{15}}\\ = \frac{{15}}{{60}} + \frac{{ - 44}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\)
Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)
P/s:Câu B tương tự nhé
Tiếp B của @Phạm Tuấn Đạt
\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)
\(B=\frac{36}{9}+\frac{13}{5}\)
\(B=4+\frac{13}{5}\)
\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)