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

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

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) \(B = \frac{\frac{3}{4} \left(\right. \frac{1}{5} - \frac{2}{7} - \frac{1}{3} + \frac{2}{7} \left.\right)}{\frac{1}{5} \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right) - \frac{1}{3} \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right)} = \frac{\frac{3}{4} \left(\right. \frac{1}{5} - \frac{1}{3} \left.\right)}{\left(\right. \frac{1}{5} - \frac{1}{3} \left.\right) \left(\right. \frac{2}{7} + \frac{1}{3} \left.\right)} = 1 \frac{11}{52}\).

) \(A = \frac{3}{5} . \&\text{nbsp}; \frac{6}{7} + \frac{3}{7} . \&\text{nbsp}; \frac{3}{5} - \frac{2}{7} . \&\text{nbsp}; \frac{3}{5}\)
\(= \frac{3}{5} \cdot \left(\right. \frac{6}{7} + \frac{3}{7} - \frac{2}{7} \left.\right) = \frac{3}{5}\)
b) \(\&\text{nbsp}; B = \left(\right. - 13 \cdot \frac{2}{5} + \frac{- 2}{9} \cdot \frac{2}{5} + \frac{2}{5} \cdot \frac{11}{9} \left.\right) \cdot \frac{5}{2}\)
\(= \left(\right. - 13 - \frac{2}{9} + \frac{11}{9} \left.\right) \cdot \frac{2}{5} \cdot \frac{5}{2} = - 13 + \left(\right. \frac{11}{9} - \frac{2}{9} \left.\right) = - 12.\)
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) \(D = \frac{4}{9} : \left(\right. \frac{1}{15} - \frac{10}{15} \left.\right) + \frac{4}{9} : \left(\right. \frac{2}{22} - \frac{5}{22} \left.\right)\)
\(= \frac{4}{9} : \frac{- 3}{5} + \frac{4}{9} : \frac{- 3}{22} = \frac{4}{9} \cdot \frac{- 5}{3} + \frac{4}{9} . \&\text{nbsp}; \frac{- 22}{3}\)
\(= \frac{4}{9} \cdot \left(\right. \frac{- 5}{3} + \frac{- 22}{3} \left.\right) = \frac{4}{9} . \&\text{nbsp}; \frac{- 27}{3} = - 4.\)