Dọc theo chiều dài, ta trồng được:

\(5.5 : \frac{1}{4} = 22\) (khóm hoa)

Dọc theo chiều rộng, ta trồng được:

3,75:\(\frac14\) ​=10 (khóm hoa)

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

\(\left[\right.\left(\right.22+10\left.\right).2\left]\right.-4=30\) (khóm hoa)

Đáp số: 30 khóm hoa

Dọc theo chiều dài, ta trồng được:

\(5.5 : \frac{1}{4} = 22\) (khóm hoa)

Dọc theo chiều rộng, ta trồng được:

3,75:\(\frac14\) ​=10 (khóm hoa)

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

\(\left[\right.\left(\right.22+10\left.\right).2\left]\right.-4=30\) (khóm hoa)

Đáp số: 30 khóm hoa

a) \(\frac{1}{5}+\frac{4}{5}:x=0,75\)

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

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

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

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

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) \(- \frac{3}{4} \cdot 5 \frac{3}{13} - 0 , 75 \cdot \frac{36}{13} = - \frac{3}{4} \cdot 5 \frac{3}{13} - \frac{3}{4} \cdot \frac{36}{13}\)
\(= - \frac{3}{4} \left(\right. 5 \frac{3}{13} + \frac{36}{13} \left.\right)\)
=−\(\frac34.8\)

=-6

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

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

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

\(x=\frac{1}{10}\)

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) \(\left(\right. x : 2 \frac{1}{3} \left.\right) \cdot \frac{1}{7} = \frac{- 3}{8}\)
(\(x\) :\(2\frac13\) )

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

\(x = \frac{- 3}{8} : \frac{3}{49}\)
\(x = \frac{- 49}{8} = - 6 \frac{1}{8}\)

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=\(\frac{-59}{40}\) \(:\frac{-4}{7}\)

\(x=\frac{413}{160}\) ​

​
​

a) \(\frac{1}{2} - \left(\right. x + \frac{1}{3} \left.\right) = \frac{5}{6}\)
x+\(\frac13\) ​=\(\frac12\) ​−\(\frac56\) ​

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

\(x = - \frac{2}{3}\)

b) \(\left(\right. \frac{3}{8} - \frac{1}{5} \left.\right) + \left(\right. \frac{5}{8} - x \left.\right) = \frac{1}{5}\)

\(\frac38\) ​+\(\frac58\) ​−\(\frac15\) ​−x=\(\frac15\) ​

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

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

a) \(- 1 , 62 + \frac{2}{5} + x = 7\)

 \(x=7-\frac{2}{5}+1,62\)

\(x\) =8,22

b) \(4 \frac{3}{5} - x = \frac{- 1}{5} + \frac{1}{2}\)

  \(x=4\frac{3}{5}+\frac{1}{5}-\frac{1}{2}\)

x=\(4\frac{3}{10}\)

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

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

x=\(\frac{41}{35}\)

d) \(\frac{5}{7} - \frac{1}{13} + 0 , 25 = 3 \frac{1}{2} - x\)

x= \(3\frac12-\frac57\) \(+\frac{1}{13}-0,25\)

x=\(3\frac{223}{364}\) ​
​

​

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) \(D = \frac{4}{9} : \left(\right. \frac{1}{15} - \frac{10}{15} \left.\right) + \frac{4}{9} : \left(\right. \frac{2}{22} - \frac{5}{22} \left.\right)\)
\(=\frac{4}{9}:\frac{- 3}{5}+\frac{4}{9}:\frac{- 3}{22}=\frac{4}{9}\cdot\frac{- 5}{3}+\frac{4}{9}.\frac{- 22}{3}\)
\(=\frac{4}{9}\cdot\left(\right.\frac{- 5}{3}+\frac{- 22}{3}\left.\right)=\frac{4}{9}.\frac{- 27}{3}=-4.\)
\(\)
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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) \(Q = \left(\right. 5 - 6 - 2 \left.\right) + \left(\right. - \frac{3}{4} - \frac{7}{4} + \frac{5}{4} \left.\right) + \left(\right. \frac{1}{5} + \frac{8}{5} - \frac{16}{5} \left.\right) = - \left(\right. 3 + \frac{5}{4} + \frac{7}{5} \left.\right)\) \(=-\frac{113}{20}\).