a) \(0 , \left(\right. 3 \left.\right) + 3 \frac{1}{2} + 0 , 4 \left(\right. 2 \left.\right)\).

-> Ta đưa \(0 , \left(\right. 3 \left.\right)\) và \(0 , 4 \left(\right. 2 \left.\right)\) về phân số như sau:

+ Đặt \(x = 0 , \left(\right. 3 \left.\right)\) thì \(10 x = 3 , \left(\right. 3 \left.\right) = 3 + 0 , \left(\right. 3 \left.\right) = 3 + x\).

Suy ra \(9 x = 3\) hay \(x = \frac{1}{3} = 0 , \left(\right. 3 \left.\right)\).

+ Ta có \(0 , 4 \left(\right. 2 \left.\right) = 0 , 4 + 0 , 0 \left(\right. 2 \left.\right)\)

Đặt \(y = 0 , 0 \left(\right. 2 \left.\right)\) thì \(100 y = 2 , \left(\right. 2 \left.\right) = 2 + 10 y\)

Suy ra \(90 y = 2\) hay \(y = 0 , 0 \left(\right. 2 \left.\right) = \frac{1}{45}\).

Do đó, \(0 , 4 \left(\right. 2 \left.\right) = 0 , 4 + 0 , 0 \left(\right. 2 \left.\right) = \frac{19}{45}\).

-> Quay trở lại bài toán: \(0 , \left(\right. 3 \left.\right) + 3 \frac{1}{2} + 0 , 4 \left(\right. 2 \left.\right) = \frac{1}{3} + \frac{7}{2} + \frac{19}{45} = \frac{383}{90} .\)

b) \(\frac{4}{9} + 1 , 2 \left(\right. 31 \left.\right) - \&\text{nbsp}; 0 , \left(\right. 13 \left.\right)\).

 Ta đưa \(1 , 2 \left(\right. 31 \left.\right)\) và \(0 , \left(\right. 13 \left.\right)\) về phân số như sau:

Đặt \(x = 0 , \left(\right. 01 \left.\right)\) thì \(100 x = 1 , \left(\right. 01 \left.\right) = 1 + x\).

Suy ra \(99 x = 1\) hay \(x = \frac{1}{99} = 0 , \left(\right. 01 \left.\right)\).

+ Tính \(1 , 2 \left(\right. 31 \left.\right)\):

Xét \(0,\left(\right.31\left.\right)=0,\left(\right.01\left.\right);31=31.\frac{1}{99}=\frac{31}{99}\).

Vậy \(1 , 2 \left(\right. 31 \left.\right) = 1 + 0 , 2 + 0 , 0 \left(\right. 31 \left.\right) = 1 + \frac{1}{5} + \frac{31}{990} = \frac{1219}{990}\).

+ Tính \(0 , \left(\right. 13 \left.\right) = 13.0 , 0 \left(\right. 1 \left.\right) = 13. \frac{1}{99} = \frac{13}{99}\).

 Quay trở lại bài toán: \(\frac{4}{9} + 1 , 2 \left(\right. 31 \left.\right) - \&\text{nbsp}; 0 , \left(\right. 13 \left.\right) = \frac{4}{9} + \frac{1219}{990} - \frac{13}{99} = \frac{139}{90}\).

a) \(m = \sqrt{25 + 9}\) và \(n = \sqrt{25} + \sqrt{9}\).

Ta có \(m = \sqrt{34}\) và \(n = 5 + 3 = 8 = \sqrt{64}\).

Mà \(34 < 64\) nên \(m < n\).

b) \(y = \sqrt{49 - 16}\) và \(z = \sqrt{81} - \sqrt{9}\).

Ta có \(y = \sqrt{49 - 16} = \sqrt{33}\) và \(z=9-3=6=\sqrt{36}\).

Mà \(33 < 36\) nên \(y < z\)

a) A = \(\sqrt{36} . \left(\right. 3 \sqrt{4} - \sqrt{\frac{1}{9}} \left.\right) + 2\)

= \(6. \left(\right. 3.2 - \frac{1}{3} \left.\right) + 2\)

= \(36 - 2 + 2 = 36.\)

b) B = \(\sqrt{\frac{1}{9} + \frac{1}{16}}\)

= \(\sqrt{\frac{9 + 16}{9.16}}\)

= \(\sqrt{\frac{5^{2}}{3^{2} . 4^{2}}}\)

= \(\frac{5}{12}\).

c) C = \(\left(\right.\sqrt{\frac{1}{9}}+\sqrt{\frac{25}{36}}-\sqrt{\frac{49}{81}}\left.\right):\sqrt{\frac{441}{324}}\)

= \(\left(\right. \frac{1}{3} + \frac{5}{6} - \frac{7}{9} \left.\right) : \sqrt{\frac{2 1^{2}}{1 8^{2}}}\)

= \(\frac{7}{18} : \frac{7}{6}\)

= \(\frac{1}{3}\).

d) \(\sqrt{\left(\frac{- 2}{5}\right)^2}+\sqrt{1 , 44}-\sqrt{256}\)

= \(\frac{2}{5} + 1 , 2 - 16\)

= \(- \frac{72}{5}\).

a)\(\left(\right. \frac{1}{2} + 1 , 5 \left.\right) \cdot x = \frac{1}{5}\)

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

\(x = \frac{1}{5} : 2\)

\(x=\frac{1}{10}\)
b) \(\left(\right. - 1 \frac{3}{5} + x \left.\right) : \frac{12}{13} = 2 \frac{1}{6}\)

\(- 1 \frac{3}{5} + x = \frac{13}{6} \cdot \frac{12}{13}\)
\(x = 2 + 1 \frac{3}{5}\)

\(x=3\frac{3}{5}\)
c) \(\left(\right. x : 2 \frac{1}{3} \left.\right) \cdot \frac{1}{7} = \frac{- 3}{8}\)

\(x \cdot \frac{3}{7} \cdot \frac{1}{7} = \frac{- 3}{8}\)

\(x = \frac{- 3}{8} : \frac{3}{49}\)
\(x = \frac{- 49}{8} = - 6 \frac{1}{8}\)
d) \(\frac{- 4}{7} \cdot x + \frac{7}{5} = \frac{1}{8} : \left(\right. - 1 \frac{2}{3} \left.\right)\)

\(\frac{- 4}{7} x + \frac{7}{5} = \frac{1}{8} \cdot \frac{- 3}{5}\)
\(- \frac{4}{7} x = \frac{- 3}{40} - \frac{7}{5} x = \frac{- 59}{40} : \frac{- 4}{7} = \frac{413}{160} = 2 \frac{93}{160}\)

a) \(x + \frac{1}{3} = \frac{1}{2} - \frac{5}{6}\)

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

\(x = - \frac{2}{3}\);
b) \(\frac{3}{8} + \frac{5}{8} - \frac{1}{5} - x = \frac{1}{5}\)

\(x = 1 - \frac{1}{5} - \frac{1}{5}\)

\(x = \frac{3}{5}\).

a) \(x = 7 - \frac{2}{5} + 1 , 62 = 8 , 22\)
b) \(x = 4 \frac{3}{5} + \frac{1}{5} - \frac{1}{2} = 4 \frac{3}{10}\)
c) \(2 x - x = \frac{3}{5} + \frac{4}{7}\)
\(x = \frac{41}{35}\)
d) \(x = 3 \frac{1}{2} - \frac{5}{7} + \frac{1}{13} - 0.25\)
\(x = 2 \frac{223}{364}\)

a) \(A = \left[\right. \frac{2}{7} \left(\right. \frac{1}{4} - \frac{1}{3} \left.\right) \left]\right. : \left[\right. \frac{2}{7} \left(\right. \frac{1}{3} - \frac{2}{5} \left.\right) \left]\right. = \left(\right. \frac{1}{4} - \frac{1}{3} \left.\right) : \left(\right. \frac{1}{3} - \frac{2}{5} \left.\right) = 1 \frac{1}{4}\).
b) \(B = \frac{\frac{3}{4} \left(\right. \frac{1}{5} - \frac{2}{7} - \frac{1}{3} + \frac{2}{7} \left.\right)}{\frac{1}{5} \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right) - \frac{1}{3} \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right)} = \frac{\frac{3}{4} \left(\right. \frac{1}{5} - \frac{1}{3} \left.\right)}{\left(\right. \frac{1}{5} - \frac{1}{3} \left.\right) \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right)} = 1 \frac{11}{52}\).

a) \(A=\frac{3}{5}.\frac{6}{7}+\frac{3}{7}.\frac{3}{5}-\frac{2}{7}.\frac{3}{5}\)
\(= \frac{3}{5} \cdot \left(\right. \frac{6}{7} + \frac{3}{7} - \frac{2}{7} \left.\right) = \frac{3}{5}\)
b) \(B=\left(\right.-13\cdot\frac{2}{5}+\frac{- 2}{9}\cdot\frac{2}{5}+\frac{2}{5}\cdot\frac{11}{9}\left.\right)\cdot\frac{5}{2}\)
\(= \left(\right. - 13 - \frac{2}{9} + \frac{11}{9} \left.\right) \cdot \frac{2}{5} \cdot \frac{5}{2} = - 13 + \left(\right. \frac{11}{9} - \frac{2}{9} \left.\right) = - 12.\)
c) \(C = \left(\right. \frac{- 4}{5} + \frac{5}{7} \left.\right) \cdot \frac{3}{2} + \left(\right. \frac{- 1}{5} + \frac{2}{7} \left.\right) \cdot \frac{3}{2} = \left(\right. \frac{- 4}{5} + \frac{5}{7} + \frac{- 1}{5} + \frac{2}{7} \left.\right) \cdot \frac{3}{2} = \left(\right. \left(\right. \frac{- 4}{5} + \frac{- 1}{5} \left.\right) + \left(\right. \frac{5}{7} + \frac{2}{7} \left.\right) \left.\right) \cdot \frac{3}{2} = 0.\)
d) \(D = \frac{4}{9} : \left(\right. \frac{1}{15} - \frac{10}{15} \left.\right) + \frac{4}{9} : \left(\right. \frac{2}{22} - \frac{5}{22} \left.\right)\)
\(=\frac{4}{9}:\frac{- 3}{5}+\frac{4}{9}:\frac{- 3}{22}=\frac{4}{9}\cdot\frac{- 5}{3}+\frac{4}{9}.\frac{- 22}{3}\)
\(=\frac{4}{9}\cdot\left(\right.\frac{- 5}{3}+\frac{- 22}{3}\left.\right)=\frac{4}{9}.\frac{- 27}{3}=-4.\)

a) \(P = \frac{2}{3} + \frac{1}{4} + \frac{3}{5} - \frac{7}{45} + \frac{5}{9} + \frac{1}{12} + \frac{1}{35}\) \(= \left(\right. \frac{2}{3} + \frac{1}{4} + \frac{1}{12} \left.\right) + \left(\right. \frac{5}{9} - \frac{7}{45} \left.\right) + \frac{3}{5} + \frac{1}{35} = 1 + \frac{4}{5} + \frac{3}{5} + \frac{1}{35} = 2 \frac{1}{35}\).
b) \(Q = \left(\right. 5 - 6 - 2 \left.\right) + \left(\right. - \frac{3}{4} - \frac{7}{4} + \frac{5}{4} \left.\right) + \left(\right. \frac{1}{5} + \frac{8}{5} - \frac{16}{5} \left.\right) = - \left(\right. 3 + \frac{5}{4} + \frac{7}{5} \left.\right)\) \(=-\frac{113}{20}\).

a) \(A = \left(\right. \frac{1}{3} + \frac{2}{3} \left.\right) - \left(\right. \frac{8}{15} + \frac{7}{15} \left.\right) + \left(\right. \frac{- 1}{7} + 1 \frac{1}{7} \left.\right) = 1 - 1 + 1 = 1\);
b) \(B = \left(\right. 0.25 - 1 \frac{1}{4} \left.\right) + \left(\right. \frac{3}{5} + \frac{2}{5} \left.\right) - \frac{1}{8}\)
\(= \left(\right. \frac{1}{4} - 1 - \frac{1}{4} \left.\right) + 1 - \frac{1}{8} = \frac{- 1}{8}\).