Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{3+\frac{3}{4}-\frac{3}{19}}{\frac{5}{2}+\frac{5}{8}-\frac{5}{38}}.\frac{15+\frac{15}{34}+\frac{15}{17}-\frac{15}{144}}{2+\frac{2}{34}+\frac{2}{17}-\frac{2}{144}}+2018\)
\(=\frac{3\left(1+\frac{1}{4}-\frac{1}{19}\right)}{\frac{5}{2}\left(1+\frac{1}{4}-\frac{1}{19}\right)}.\frac{15\left(1+\frac{1}{34}+\frac{1}{17}-\frac{1}{144}\right)}{2\left(1+\frac{1}{34}+\frac{1}{17}-\frac{1}{144}\right)}+2018\)
\(=\frac{3}{\frac{5}{2}}.\frac{15}{2}+2018\)
\(=\frac{6}{5}.\frac{15}{2}+2018\)
\(=9+2018=2027\)
\(\frac{3+0,75-\frac{3}{19}}{\frac{5}{2}+0,625-\frac{5}{38}}\) . \(\frac{15+\frac{15}{17}-\frac{15}{144}+\frac{15}{34}}{2+\frac{2}{17}-\frac{2}{144}+\frac{1}{17}}\)+2018
=\(\frac{3+\frac{3}{4}-\frac{3}{19}}{\frac{5}{2}+\frac{5}{8}-\frac{5}{38}}\) . \(\frac{15+\frac{15}{17}-\frac{15}{144}+\frac{15}{34}}{2+\frac{2}{17}-\frac{2}{144}+\frac{2}{34}}\)+2018
=\(\frac{3\left(1+\frac{1}{4}-\frac{1}{19}\right)}{5\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}\) . \(\frac{15\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}{2\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}\)+2018
=\(\frac{6\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}{5\left(\frac{1}{2}+\frac{1}{8}-\frac{1}{38}\right)}\) . \(\frac{15\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}{2\left(1+\frac{1}{17}-\frac{1}{144}+\frac{1}{34}\right)}\)+2018
=\(\frac{6}{5}\) . \(\frac{15}{2}\)+2018
=9+2018
=2027
Thật ra đề không sai đâu nha các bạn chúng ta cần phân tích để tử số giống mẫu số rồi rút gọn chúng ta sẽ ra được kết quả
Chúc các bạn học tốt
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}\)
\(=\frac{12}{17}+\frac{1}{3}-\frac{32}{17}+\frac{2}{3}\)
\(=-\frac{20}{17}+\frac{1}{3}+\frac{2}{3}\)
\(=-\frac{20}{17}+1\)
\(=-\frac{3}{17}\)
\(B=16\frac{2}{7}:\left(-\frac{3}{5}\right)-28\frac{2}{7}:\left(-\frac{3}{5}\right)\)
\(=\left(\frac{114}{7}-\frac{198}{7}\right):\left(-\frac{3}{5}\right)\)
\(=\left(-12\right):\left(-\frac{3}{5}\right)\)
\(=20\)
A\(=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}\)
\(=\left(\frac{15}{34}+\frac{9}{34}\right)+\left(\frac{7}{21}+\frac{2}{3}\right)-1\frac{15}{17}\)
\(=\left(\frac{14}{17}-\frac{15}{17}\right)+1-1\)
\(=-\frac{1}{17}\)
\(B=16\frac{2}{7}:\left(-\frac{3}{5}\right)-28\frac{2}{7}:\left(-\frac{3}{5}\right)\)
\(B=-\frac{3}{5}.\left(16\frac{2}{7}-28\frac{2}{7}\right)\)
\(B=-\frac{3}{5}.\left(-12\right)\)
\(B=\frac{36}{5}\)
Hoctot
a)
\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)
c)
\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)
d)
\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)
e)
\(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)
g)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
