Giải

Chu vi mảnh vườn hình chữ nhật đó là:

(5,5+3,75).2=18,5 (m)

Số khóm hoa cần trồng ban đầu là:

18,5:\(\dfrac{1}{4}\)= 74 (khóm)

Theo đề bài: dọc theo các cạnh của mảnh vườn để trồng những khóm hoa, do đó không tính 4 khóm hoa trồng ở bốn góc vườn nên số khóm hoa cần trồng là:

74-4=70 (khóm)

Vậy cần trồng 70 khóm hoa 

a)\(\dfrac{1}{5}+\dfrac{4}{5}:x=0,75\)

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

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

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)

b)\(x+\dfrac{1}{2}=1-x\)

\(\Leftrightarrow x+x=1-\dfrac{1}{2}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4}\)

a)\(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{3}\)

b)\(2.\left(\dfrac{-3}{2}\right)^{2^{ }}-\dfrac{7}{2}\)

\(=2.\dfrac{9}{4}-\dfrac{7}{2}\)

\(=\dfrac{9}{2}-\dfrac{7}{2}\)

\(=1\)

c)\(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)

\(=-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

\(=\dfrac{3}{4}.\left(\dfrac{-68}{13}\right)-\dfrac{3}{4}.\dfrac{36}{13}\)

\(=\dfrac{3}{4}.\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)

\(=\dfrac{3}{4}.\left(-8\right)\)

\(=\dfrac{-24}{4}\)

\(=-6\)

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

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{3}{2}\right).x=\dfrac{1}{5}\)

\(\Leftrightarrow2x=\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{1}{10}\)

Vậy \(x=\dfrac{1}{10}\)

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

\(\Leftrightarrow\left(\dfrac{-8}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)

\(\Leftrightarrow\left(\dfrac{-8}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow\left(\dfrac{-8}{5}+x\right)=2\)

\(\Leftrightarrow\dfrac{-8}{5}+x=2\)

\(\Leftrightarrow x=2-\left(\dfrac{-8}{5}\right)\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

Vậy \(x=\dfrac{18}{5}\)

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

\(\Leftrightarrow\left(x:\dfrac{7}{3}\right).\dfrac{1}{7}=\dfrac{-3}{8}\)

\(\Leftrightarrow\left(x:\dfrac{7}{3}\right)=\dfrac{-3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x:\dfrac{7}{3}=\dfrac{-21}{8}\)

\(\Leftrightarrow x=\dfrac{-21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=\dfrac{-49}{8}\)

Vậy \(x=\dfrac{-49}{8}\)

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

\(\Leftrightarrow\dfrac{-4}{7}.x+\dfrac{7}{5}=\dfrac{1}{8}:\left(\dfrac{-5}{3}\right)\)

\(\Leftrightarrow\dfrac{-4}{7}.x+\dfrac{7}{5}=\dfrac{-3}{40}\)

\(\Leftrightarrow\dfrac{-4}{7}.x=\dfrac{-3}{40}-\dfrac{7}{5}\)

\(\Leftrightarrow\dfrac{-4}{7}.x=\dfrac{-59}{40}\)

\(\Leftrightarrow x=\dfrac{-59}{40}:\left(\dfrac{-4}{7}\right)\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

Vậy \(x=\dfrac{413}{160}\)

a)\(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)=\dfrac{-1}{3}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(x=\dfrac{-2}{3}\)

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

\(\Leftrightarrow\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}-\left(\dfrac{3}{8}-\dfrac{1}{5}\right)\)

\(\Leftrightarrow\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}-\dfrac{7}{40}\)

\(\Leftrightarrow\left(\dfrac{5}{8}-x\right)=\dfrac{1}{40}\)

\(\Leftrightarrow\dfrac{5}{8}-x=\dfrac{1}{40}\)

\(\Leftrightarrow x=\dfrac{3}{5}\)

Vậy \(x=\dfrac{3}{5}\)

\(-1,62+\dfrac{2}{5}+x=7\)

\(\Leftrightarrow x=7-\dfrac{2}{5}+1,62\)

\(\Leftrightarrow x=\dfrac{33}{5}+1,62\)

\(\Leftrightarrow x=\dfrac{33}{5}+\dfrac{81}{50}\)

\(\Leftrightarrow x=8,22\)

Vậy \(x=8,22\)

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

\(\Leftrightarrow\dfrac{23}{5}-x=\dfrac{-1}{5}+\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{23}{5}+\dfrac{1}{5}-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{24}{5}-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{48}{10}-\dfrac{5}{10}\)

\(\Leftrightarrow x=\dfrac{43}{10}\)

Vậy \(x=\dfrac{43}{10}\)

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

\(\Leftrightarrow2x-x=\dfrac{3}{5}+\dfrac{4}{7}\)

\(\Leftrightarrow x=\dfrac{41}{35}\)

Vậy \(x=\dfrac{41}{35}\)

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

\(\Leftrightarrow\dfrac{5}{7}-\dfrac{1}{13}+\dfrac{1}{4}=\dfrac{7}{2}-x\)

\(\Leftrightarrow x=\dfrac{7}{2}-\dfrac{5}{7}+\dfrac{1}{13}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{39}{14}+\dfrac{1}{13}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{507}{182}+\dfrac{14}{182}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{521}{182}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1042}{364}-\dfrac{91}{364}\)

\(\Leftrightarrow x=\dfrac{951}{364}\)

Vậy \(x=\dfrac{951}{364}\)

\(A=\left(\dfrac{2}{7}.\dfrac{1}{4}-\dfrac{1}{3}.\dfrac{2}{7}\right):\left(\dfrac{2}{7}.\dfrac{3}{9}-\dfrac{2}{7}.\dfrac{2}{5}\right)\)

    \(=\left(\dfrac{2}{7}.\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right):\left(\dfrac{2}{7}.\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right)\)

    \(=\left(\dfrac{2}{7}.\dfrac{-1}{12}\right):\left(\dfrac{2}{7}.\dfrac{-1}{15}\right)\)

    \(=\dfrac{-1}{12}:\dfrac{-1}{15}\)

    \(=\dfrac{5}{4}\)

\(A=\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}\)

    \(=\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{2}{7}.\dfrac{3}{5}\) 

    \(=\dfrac{3}{5}.\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right)\)

    \(=\dfrac{3}{5}.1\)

    \(=\dfrac{3}{5}\)

\(B=\left(-13.\dfrac{2}{5}+\dfrac{-2}{9}:2\dfrac{1}{2}+\dfrac{2}{5}.\dfrac{11}{9}\right).2\dfrac{1}{2}\)

    \(=\left(-13.\dfrac{2}{5}+\dfrac{-2}{9}.\dfrac{2}{5}+\dfrac{2}{5}.\dfrac{11}{9}\right).\dfrac{5}{2}\)

    \(=\left(-13+\dfrac{-2}{9}+\dfrac{11}{9}\right).\dfrac{2}{5}.\dfrac{5}{2}\)

    \(=\left(-13+\dfrac{-2}{9}+\dfrac{11}{9}\right).1\)

    \(=\left(-13+\left(\dfrac{-2}{9}+\dfrac{11}{9}\right)\right).1\)

    \(=\left(-13+1\right).1\)

    \(=-12\)

\(C=\left(\dfrac{-4}{5}+\dfrac{5}{7}\right):\dfrac{2}{3}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right):\dfrac{2}{3}\)

    \(=\left(\dfrac{-4}{5}+\dfrac{5}{7}\right).\dfrac{3}{2}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right).\dfrac{3}{2}\)

    \(=\dfrac{3}{2}.\left(\left(\dfrac{-4}{5}+\dfrac{5}{7}\right)+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right)\right)\)

    \(=\dfrac{3}{2}.\left(\left(\dfrac{-4}{5}+\left(\dfrac{-1}{5}\right)\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)\right)\)

    \(=\dfrac{3}{2}.\left(-1+1\right)\)

    \(=\dfrac{3}{2}.0=0\)

\(D=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)

    \(=\dfrac{4}{9}:\dfrac{-3}{5}+\dfrac{4}{9}:\dfrac{-3}{22}\)

    \(=\dfrac{4}{9}.\dfrac{-5}{3}+\dfrac{4}{9}.\dfrac{-22}{3}\)

    \(=\dfrac{4}{9}.\left(\dfrac{-5}{3}+\dfrac{-22}{3}\right)\)

    \(=\dfrac{4}{9}.\left(-9\right)\)

    \(=-4\)

\(P=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

    \(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}+\left(\dfrac{-7}{45}\right)+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

    \(=\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{-7}{45}+\dfrac{5}{9}\right)+\dfrac{3}{5}+\dfrac{1}{35}\)

    \(=1+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{1}{35}\)

    \(=1+\dfrac{14}{35}+\dfrac{21}{35}+\dfrac{1}{35}\)

    \(=\dfrac{49}{35}+\dfrac{21}{35}+\dfrac{1}{35}\)

    \(=\dfrac{70}{35}+\dfrac{1}{35}\)

   \(=\dfrac{71}{35}\)

\(Q=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)

    \(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)

    \(=5+\left(\dfrac{-3}{4}\right)+\dfrac{1}{5}+\left(-6\right)+\left(\dfrac{-7}{4}\right)+\dfrac{8}{5}+\left(-2\right)+\dfrac{5}{4}+\left(\dfrac{-16}{5}\right)\)

    \(=\left(5+\left(-6\right)+\left(-2\right)\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}+\left(\dfrac{-16}{5}\right)\right)+\left(\left(\dfrac{-3}{4}\right)+\left(\dfrac{-7}{4}\right)+\dfrac{5}{4}\right)\)

    \(=-3+\left(\dfrac{-7}{5}\right)+\left(\dfrac{-5}{4}\right)\)

    \(=\dfrac{-60}{20}+\left(\dfrac{-28}{20}\right)+\left(\dfrac{-25}{20}\right)\)

    \(=\dfrac{-88}{20}+\left(\dfrac{-25}{20}\right)\)

    \(=\dfrac{-113}{20}\)

\(A=\left(\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}\right)+\left(\dfrac{2}{3}+\dfrac{-7}{15}+1\dfrac{1}{7}\right)\)

    \(=\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}+\dfrac{2}{3}+\dfrac{-7}{15}+1\dfrac{1}{7}\)

    \(=\dfrac{1}{3}+\left(\dfrac{-8}{15}\right)+\left(\dfrac{-1}{7}\right)+\dfrac{2}{3}+\left(\dfrac{-7}{15}\right)+\dfrac{8}{7}\)

    \(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\left(\dfrac{-8}{15}\right)+\left(\dfrac{-7}{15}\right)\right)+\left(\left(\dfrac{-1}{7}\right)+\dfrac{8}{7}\right)\)

    \(=1+\left(-1\right)+1\)

    \(=1\)

\(B=0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

    \(=\dfrac{1}{4}+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+\dfrac{5}{4}\right)\)

    \(=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}\)

    \(=\dfrac{1}{4}+\dfrac{3}{5}+\left(\dfrac{-1}{8}\right)+\dfrac{2}{5}+\left(\dfrac{-5}{4}\right)\)

    \(=\left(\dfrac{1}{4}+\left(\dfrac{-5}{4}\right)\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{-1}{8}\right)\)

    \(=-1+1+\left(\dfrac{-1}{8}\right)\)

    \(=\dfrac{-1}{8}\)