Cách nêu tính chất đặc trưng:

A=\(\left\{x/\left(x^2+2x-3\right)\left(x^2-13x+42\right)\right\}\)

B=\(\left\{\frac{2x+1}{2^{x+1}},x\in N,0\le x\le4\right\}\)

Bài 1:

a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)

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

\(=4\frac{5}{7}-1\frac{3}{4}\)

\(=\frac{33}{7}-\frac{7}{4}\)

\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)

b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)

\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)

\(=4\frac{5}{9}-2\frac{3}{4}\)

\(=\frac{41}{9}-\frac{11}{4}\)

\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)

c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)

\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)

\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)

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

\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)

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

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

a,  \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)

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

\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)

\(A=\frac{-7}{20}+\frac{1}{2}\)

\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)

b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)

\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)

\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)

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

\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)

\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)

c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)

\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)

\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)

\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)

\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)

\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)

d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)

\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)

A=3/20

B=-1/4

C=-9/95

D=-59/111

a) \(A=\left\{x\in N|0\le x\le4\right\}\)

b) \(B=\left\{x\in N|x=4k;0\le k\le4;k\in N\right\}\)

c) \(C=\left\{x\in Z|x=\left(-3\right)^k;1\le k\le4;k\in N\right\}\)

d) \(D=\left\{x\in N|x=k^2;k=3a;1\le a\le4;a\in N\right\}\)

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

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

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

\(=14\)