a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)

\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)

\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)

b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)

c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)

a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)

b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10

a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)

b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)

c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)

d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)

e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)

kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi

trời ơi bài dễ thế này tự làm đi còn hỏi

Đã biết thì đã không hỏi. Đồ kazuto kirigaya xấu tính!

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) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)

b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) ^{_{(Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)}}

c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)

\(=\frac{-3}{5}.10\) \(=-6\)

d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)

\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)

a)    

b)   

c)   

d)  

a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)

=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)

=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)

=\(-\frac{11}{4}\cdot\frac{8}{33}\)

=\(-\frac{2}{3}\)

b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)

=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)

=\(\frac{-1}{11}\cdot55=-5\)

c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)

=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)

=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)

=\(-\frac{2}{3}\cdot\frac{7}{5}\)

=\(\frac{-14}{15}\)

d) chưa nghĩ ra nhé

e) bạn chép sai đề bài rồi

mk mới kiểm tra 45 phút nên biết

đề bài nè

\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)

=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)

=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)

=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)

=\(\frac{1.31}{2.30}\)

=\(\frac{31}{60}\)

a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong                                                                         b)bn tu lam nhe                                                                                                                                             c)dat thua so chung                                                                                                                                       d)tinh trong ngoac ra rui nhan vs                                                                                                                       e) mk bo tay 

thanks mấy pn nha

Bài 1:

a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)

\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)

\(=\frac{2+3}{12}=\frac{5}{12}\)

b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{-7}{10}\)

\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)

Bài 2:

a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)

\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)

\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)

\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)

\(\Leftrightarrow-2x=-6-1=-7\)

hay \(x=\frac{7}{2}\)

Vậy: \(x=\frac{7}{2}\)

Bài 1:

a) Ta có: 

b) Ta có: 

Bài 2:

a) Ta có: 

Vậy: 

b) Ta có: 

hay 

Vậy: 

Những câu từ D trở đi là các câu riêng biệt ak bạn

\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)

\(B=3-{2\over3}+{3\over5}(-{10\over9}-{25\over3})-{5\over6}\)\(={7\over3}+{3\over5}(-{10\over9}-{75\over9})-{5\over6}\)\(={14\over6}-{5\over6}+{3\over5}({-85\over9})\)\(={3\over2}-{17\over3}={9\over6}-{34\over6}={-25\over6}\)