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minh vu
Shiratori Hime
KYAN Gaming

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Câu trả lời:

ĐỀ 1:

Bài 1:

a) \(\dfrac{2}{3}-\dfrac{5}{7}\cdot\dfrac{14}{25}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}\)

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

c) \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}=\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{6}{5}=\dfrac{-1}{20}\)

Bài 2:

a) \(x+\dfrac{1}{2}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}-\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)

b) \(\dfrac{4}{5}\cdot x=\dfrac{4}{7}\Leftrightarrow x=\dfrac{4}{7}:\dfrac{4}{5}\Leftrightarrow x=\dfrac{4}{7}\cdot\dfrac{5}{4}\Leftrightarrow x=\dfrac{5}{7}\)

c) \(8x=7,8\cdot x+25\)

\(\Leftrightarrow8x-7,8x=25\)

\(\Leftrightarrow0,2x=25\)

\(\Leftrightarrow x=125\)

ĐỀ 2:

Bài 1:

a) \(\dfrac{7\cdot9-14}{3-17}=\dfrac{63-14}{-14}=\dfrac{49}{-14}=-\dfrac{7}{2}\)

b) \(0,25\cdot2\dfrac{1}{3}\cdot30\cdot0,5\cdot\dfrac{8}{45}=\dfrac{14}{9}\)

c) \(\dfrac{9}{23}\cdot\dfrac{5}{8}+\dfrac{9}{23}\cdot\dfrac{3}{8}-\dfrac{9}{23}=\dfrac{9}{23}\cdot\left(\dfrac{5+3-1}{8}\right)=\dfrac{9}{23}\cdot\dfrac{7}{8}=\dfrac{63}{184}\)

Bài 2:

a) \(\dfrac{1}{2}-\left(\dfrac{2}{3}\cdot x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{3}.x-\dfrac{1}{3}=-\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{2}{3}.x=\dfrac{1}{6}\)

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

b) \(\dfrac{3}{x+5}=15\%\)

\(\Leftrightarrow\dfrac{3}{x+5}=\dfrac{3}{20}\Leftrightarrow x+5=20\Leftrightarrow x=15\)

Bài 3:

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\)

\(A=\dfrac{2-1}{1\cdot2}+\dfrac{3-2}{2\cdot3}+\dfrac{4-3}{3\cdot4}+...+\dfrac{50-49}{49\cdot50}\)

\(A=\dfrac{2}{1\cdot2}-\dfrac{1}{1\cdot2}+\dfrac{3}{2\cdot3}-\dfrac{2}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{3}{3\cdot4}+...+\dfrac{50}{49\cdot50}-\dfrac{49}{49\cdot50}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=1-\dfrac{1}{50}=\dfrac{49}{50}\)