Tiền lãi một tháng là: \(\left(\right. 2 062 400 - 2 000 000 \left.\right) : 6 = 10 400\) (đồng).

Lãi suất hàng tháng là: \(\frac{10 400.100 \%}{2 000 000} = 0 , 52 \%\).

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

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) \(A = x^{2} - 2 x + 3\) khi \(\mid x \mid = 0 , 5\).

Ta có \(\mid x \mid = 0 , 5\) thì \(x = 0 , 5\) hoặc \(x = - 0 , 5\).

+ Với \(x = 0 , 5\) ta có \(A = 0 , 5^{2} - 2.0 , 5 + 3 = 2 , 25\).

+ Với \(x = - 0 , 5\) ta có \(A = \left(\right. - 0 , 5 \left.\right)^{2} - 2. \left(\right. - 0 , 5 \left.\right) + 3 = 4 , 25\).

b) \(B = x - 3 + \mid 1 - 3 x \mid\) khi \(\mid x \mid = \frac{1}{3}\).

Ta có \(\mid x \mid = \frac{1}{3}\) thì \(x = \frac{1}{3}\) hoặc \(x = - \frac{1}{3}\).

+ Với \(x = \frac{1}{3}\) ta có \(B = \frac{1}{3} - 3 + \mid 1 - 3. \frac{1}{3} \mid = - \frac{8}{3}\).

+ Với \(x = - \frac{1}{3}\) ta có \(B = - \frac{1}{3} - 3 + \mid 1 - 3. \left(\right. - \frac{1}{3} \left.\right) \mid = - \frac{1}{3} - 3 + 2 = - \frac{4}{3}\).

a) \(A = x^{2} - 2 x + 3\) khi \(\mid x \mid = 0 , 5\).

Ta có \(\mid x \mid = 0 , 5\) thì \(x = 0 , 5\) hoặc \(x = - 0 , 5\).

+ Với \(x = 0 , 5\) ta có \(A = 0 , 5^{2} - 2.0 , 5 + 3 = 2 , 25\).

+ Với \(x = - 0 , 5\) ta có \(A = \left(\right. - 0 , 5 \left.\right)^{2} - 2. \left(\right. - 0 , 5 \left.\right) + 3 = 4 , 25\).

b) \(B = x - 3 + \mid 1 - 3 x \mid\) khi \(\mid x \mid = \frac{1}{3}\).

Ta có \(\mid x \mid = \frac{1}{3}\) thì \(x = \frac{1}{3}\) hoặc \(x = - \frac{1}{3}\).

+ Với \(x = \frac{1}{3}\) ta có \(B = \frac{1}{3} - 3 + \mid 1 - 3. \frac{1}{3} \mid = - \frac{8}{3}\).

+ Với \(x = - \frac{1}{3}\) ta có \(B = - \frac{1}{3} - 3 + \mid 1 - 3. \left(\right. - \frac{1}{3} \left.\right) \mid = - \frac{1}{3} - 3 + 2 = - \frac{4}{3}\).

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 \(z = 9 - 3 \&\text{nbsp}; = 6 \&\text{nbsp}; = \sqrt{36}\).

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 = \(\left(\right. \sqrt{\frac{1}{9}} \&\text{nbsp}; + \sqrt{\frac{25}{36}} \&\text{nbsp}; - \sqrt{\frac{49}{81}} \left.\right) : \sqrt{\frac{441}{324}}\)

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

= \(\frac{7}{18} : \frac{7}{6}\)

= \(\frac{1}{3}\).

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

= \(\frac{2}{5} + 1 , 2 - 16\)

= \(- \frac{72}{5}\).

Dọc theo chiều dài, ta trồng được:

\(5.5 : \frac{1}{4} = 22\) (khóm hoa)

Dọc theo chiều rộng, ta trồng được:

\(3 , 75 : \frac{1}{4} = 15\) (khóm hoa)

Như vậy, số khóm hoa trồng được dọc theo hai cạnh của mảnh vườn là:

\(\left[\right. \left(\right. 22 + 15 \left.\right) . 2 \left]\right. - 4 = 70\) (khóm hoa)

(Chú thích, ta phải trừ đi 4 khóm hoa do 4 khóm hoa ở 4 góc hình chữ nhật được đếm 2 lần)

Đáp số: 70 khóm hoa

a) \(\&\text{nbsp}; \frac{1}{5} + \frac{4}{5} : x = 0 , 75\)

\(\&\text{nbsp}; \frac{1}{5} + \frac{4}{5} : x = \frac{3}{4}\)

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

\(\frac{4}{5} : x = \frac{11}{20}\)

\(x = \frac{16}{11}\);

b) \(x + \frac{1}{2} = 1 - x\)

 \(2 x = 1 - \frac{1}{2}\)

 \(2 x = \frac{1}{2}\)

 \(x = \frac{1}{4}\).

a) \(\frac{2}{3} \cdot \frac{5}{4} - \frac{3}{4} \cdot \frac{2}{3} = \frac{2}{3} \cdot \left(\right. \frac{5}{4} - \frac{3}{4} \left.\right) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}\);
b) \(2 \cdot \left(\left(\right. \frac{- 3}{2} \left.\right)\right)^{2} - \frac{7}{2} = 2 \cdot \frac{9}{4} - \frac{7}{2} = \frac{9}{2} - \frac{7}{2} = 1\);
c) \(- \frac{3}{4} \cdot 5 \frac{3}{13} - 0 , 75 \cdot \frac{36}{13} = - \frac{3}{4} \cdot 5 \frac{3}{13} - \frac{3}{4} \cdot \frac{36}{13}\)
\(= - \frac{3}{4} \left(\right. 5 \frac{3}{13} + \frac{36}{13} \left.\right)\)
\(= - \frac{3}{4} \cdot 8 = - 6\).

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

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

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

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

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

\(\&\text{nbsp}; x = 3 \frac{3}{5}\)
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) \(\frac{- 4}{7} \cdot x + \frac{7}{5} = \frac{1}{8} : \left(\right. - 1 \frac{2}{3} \left.\right)\)

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