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1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)
\(=\left(\dfrac{1}{10}+\dfrac{9}{10}\right)+\left(\dfrac{2}{10}+\dfrac{8}{10}\right)+\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{4}{10}+\dfrac{6}{10}\right)+\dfrac{5}{10}\)
\(=1+1+1+1+\dfrac{5}{10}\)
\(=4+\dfrac{5}{10}\)
\(=\dfrac{45}{10}\)
\(13,25:0,5+13,25:0,25+13,25:0,125+13,25\times6\)
\(=13,25:\dfrac{1}{2}+13,25:\dfrac{1}{4}+13,25:\dfrac{1}{8}+13,25\times6\)
\(=13,25\times2+13,25\times4+13,25\times8+13,25\times6\)
\(=13,25\times\left(2+4+8+6\right)\)
\(=13,25\times20\)
\(=265\)
2,5 : 0,25 + \(\dfrac{1}{8}\) : 0,125 - \(\dfrac{1}{10}\) : 0,1
= 2,5 x 4 + \(\dfrac{1}{8}\) x 8 - \(\dfrac{1}{10}\) x 10
= 10 + 1 - 1
= 10
\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)
\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)
\(=\dfrac{3}{4}-\dfrac{5}{12}\)
\(=\dfrac{4}{12}=\dfrac{1}{3}\)
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)
\(=\frac{1}{2}\times2-\frac{1}{4}\times4+\frac{1}{8}\times8-\frac{1}{10}\times10\)
\(=1-1+1-1\)
\(=0\)