b)  \(\frac{4}{10}+\frac{-2}{9}+\frac{-3}{-5}+\frac{21}{-27}+\frac{-10}{20}\)

\(=\frac{2}{5}-\frac{2}{9}+\frac{3}{5}-\frac{7}{9}-\frac{1}{2}\)

\(=\left(\frac{2}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}+\frac{7}{9}\right)-\frac{1}{2}\)

\(=1-1-\frac{1}{2}=-\frac{1}{2}\)

a)  \(-\frac{5}{2}+\frac{1}{7}+\frac{6}{7}+\frac{1}{2}+\frac{3}{4}\)

\(=\left(\frac{-5}{2}+\frac{1}{2}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\frac{3}{4}\)

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

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

\(=-\frac{1}{4}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5=1+1+0,5=2,5\)

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)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)

\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)

\(=\left(-10\right):\left(-\frac{5}{7}\right)\)

\(=14\)

a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

 b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{17875}{13299}+\frac{-11625}{13299}\)

\(=\frac{6250}{13299}\)

(Hình như câu b hơi sai)

\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

\(b)\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{6250}{13299}\)

Sao câu b) kq lại lớn vậy nhỉ , đề bài sai ? , hay mình làm sai ?

P1 có 2 thừa số âm nhân với nhau nên P1 >0

P2 có 3 thừa số âm nhân với nhau nên P2 <0

P3 vì trong dấu ba chấm có thừa số 0/10 nên P3 =0

Vậy P2 < P3 < P1

KO khó lắm đâu. Mong bạn hiểu để bài sau tương tự thì làm được.

CHúc bạn học tốt.

a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)

Phần b T²

c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)

cảm ơn bạn nha

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)\(\frac{3}{7}\cdot19\frac{1}{3}-\frac{3}{7}\cdot33\frac{1}{3}\)

\(=\frac{3}{7}\left(19\frac{1}{3}-33\frac{1}{3}\right)=\frac{3}{7}\cdot\left(-14\right)=-6\)

c) \(9\left(-\frac{1}{3}\right)^3+\frac{1}{3}=9\cdot\left(\frac{-1}{27}\right)+\frac{1}{3}=-\frac{1}{3}+\frac{1}{3}=0\)

d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)=-\frac{7}{5}:\left(15\frac{1}{4}-25\frac{1}{4}\right)=-\frac{7}{5}:\left(-10\right)=14\)

a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b) \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)

\(=\frac{3}{7}.\frac{58}{3}-\frac{3}{7}.\frac{100}{3}\)

\(=\frac{3}{7}.\left(\frac{58}{3}-\frac{100}{3}\right)\)

\(=\frac{3}{7}.\left(-14\right)\)

\(=-6\)

c) \(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)

\(=9.\left(-\frac{1}{27}\right)+\frac{1}{3}\)

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

\(=0\)

d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)

\(=\frac{61}{4}:\left(-\frac{5}{7}\right)-\frac{101}{4}:\left(-\frac{5}{7}\right)\)

\(=\left(\frac{61}{4}-\frac{101}{4}\right):\left(-\frac{5}{7}\right)\)

\(=\left(-10\right):\left(-\frac{5}{7}\right)\)

\(=14\)

 ^...^  ^_^

\(\left(5-\frac{2}{3}+\frac{3}{7}\right):\left(24-25+\frac{4}{21}-\frac{8}{21}\right)\)

\(\left(\frac{5.21-14+9}{21}\right):\left(\frac{-21-4}{21}\right)\)=\(\left(\frac{5.21-5}{21}\right).\left(\frac{21}{-25}\right)\)=\(\frac{5\left(21-1\right)}{\left(-5\right).5}=\frac{20}{-5}\)

=-4

-0.9615384315