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nana
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bấm máy tính ra kết quả là 983/504

Trần Thị Đào
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Nguyễn Trần Thành Đạt
9 tháng 2 2017 lúc 13:29

\(\left[2\frac{11}{25}-0,84.\left(6\frac{8}{9}:2\frac{7}{12}-\frac{5}{12}.4\frac{4}{35}\right)\right]:\left[7,605:7\frac{1}{2}+3,086\right]\\ < =>\left[\frac{61}{25}-\frac{21}{25}.\left(\frac{62}{9}:\frac{31}{12}-\frac{5}{12}.\frac{144}{35}\right)\right]:\left[7,605:7,5+3,086\right]\\ < =>\left[\frac{61}{25}-\frac{21}{25}.\left(\frac{8}{3}-\frac{12}{7}\right)\right]:\left[1,014+3,086\right]\\ < =>\left[\frac{61}{25}-\frac{21}{25}.\frac{20}{21}\right]:4,1\\ < =>\left[\frac{61}{25}-\frac{4}{5}\right]:4,1\\ < =>\frac{41}{25}:4,1=0,4\)

Nguyễn Châu Mỹ Linh
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Đỗ Thị Minh Anh
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phunganhtuyet
23 tháng 4 2019 lúc 20:17

a,\(\frac{4}{9}.\frac{2}{6}=\frac{4}{27}\)

b,\(1\frac{1}{3}.\left(0,5\right)+\left(\frac{8}{15}-\frac{19}{30}\right):\frac{6}{15}\)

=\(\frac{4}{3}.\frac{1}{2}+\left(\frac{16}{30}-\frac{19}{30}\right).\frac{15}{6}\)

=\(\frac{2}{3}+\frac{-1}{10}.\frac{15}{6}\)

=\(\frac{2}{3}+\frac{-1}{4}\)

=\(\frac{8}{12}+\frac{-3}{12}=\frac{5}{12}\)

bài2

a,\(\left(\frac{2}{7}.x+\frac{3}{7}\right):2\frac{1}{5}-\frac{3}{7}=1\)

=>\(\left(\frac{2}{7}.x+\frac{3}{7}\right):\frac{11}{5}=1+\frac{3}{7}=\frac{10}{7}\)

=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{10}{7}.\frac{11}{5}\)

=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{22}{7}\)

=>\(\frac{2}{7}.x=\frac{22}{7}-\frac{3}{7}=\frac{19}{7}\)

=>\(x=\frac{19}{7}:\frac{2}{7}=\frac{19}{7}.\frac{7}{2}=\frac{19}{2}\)

vậy x\(=\frac{19}{2}\)

pham thi hoa
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Tiểu Thư Bánh Bèo
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Quoc Tran Anh Le
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Hà Quang Minh
19 tháng 9 2023 lúc 20:19

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)

\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)

Phí Quỳnh Anh
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Phương Trình Hai Ẩn
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