1)\(\frac{1}{7}\).(\(\frac{1}{3}\)+\(\frac{1}{2}\)-1)

=\(\frac{1}{7}\).\(\frac{-1}{6}\)

=\(\frac{-1}{42}\)

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

=\(\frac{3}{5}\).2=\(\frac{6}{5}\)

3)=21.\(\frac{1}{7}\)-21.\(\frac{1}{5}\)+21.\(\frac{19}{21}\)

=3-\(\frac{21}{5}\)+19

=\(\frac{89}{5}\)

cảm ơn bạn nhé

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)     \(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)

\(=\left(1\frac{3}{19}-\frac{3}{19}\right)+\left(\frac{8}{21}+\frac{13}{21}\right)+0.5\)

\(=1+1+0.5=2.5\)

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

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

\(=0:\frac{3}{7}=0\)

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

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

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

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

\(=14\)

 ^...^  ^_^

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

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

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

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

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

a)|-10|:(-2):(-5)+(-3)²

    =1+9

     =10

b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)

   =[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]

   =(-1)+(-1)+(-1)+...+(-1)

Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)

        Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)

c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)

\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)

\(=\frac{3}{4}\)

d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)

\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)

\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)

\(=-\frac{1099}{1955}\)

e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)

\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)

\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)

\(=-\frac{11}{4}\)

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

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

=-4

-0.9615384315

Đặt mỗi số số hạng của M là A;B

Ta quy đồng tử và mẫu của A bằng cách nhân cả tử và mẫu với 3.7.13;rồi dùng tính chất phân phối bỏ ngoặc ra.Tính toán bình thường.Khi đó A=1/7 

Tương tự với B ta quy đồng lên với 11.17.19;bỏ ngoặc đi và tính bình thường ;B=2/7

=>M=1/7+2/7-136/7=3/7-136/7=-19