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tính bằng cách thuận tiện nhất
a x 3,7 + a x 5,3 + a = 45
a) \(\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(-\frac{1}{3}-1\frac{1}{15}\right)\)
\(=\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}-\frac{2}{6}-1+\frac{1}{15}\right)\)
\(=\frac{3}{5}:\left(-\frac{1}{2}-1\right)\)
\(=\frac{3}{5}:\left(-\frac{3}{2}\right)\)
\(=-\frac{2}{5}\)
b) \(\left(-\frac{3}{4}+\frac{5}{13}\right):\frac{2}{7}-\left(2\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)
\(=\left(-\frac{3}{4}+\frac{5}{13}-2+\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)
\(=\left(-\frac{1}{2}+1-2\right):\frac{2}{7}\)
\(=\left(-\frac{1}{2}-1\right):\frac{2}{7}\)
\(=-\frac{3}{2}:\frac{2}{7}\)
\(=-\frac{21}{4}\)
a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(-\frac{1}{10}< =x< =\frac{3}{5}\)
\(\frac{-4}{9}< x< =\frac{2}{3}\)
a) \(\frac{15}{16}\cdot\frac{4}{3}-\frac{1}{2}:\frac{5}{4}+3\)
\(=\frac{5}{4}-\frac{2}{5}+3\)
\(=\frac{25}{20}-\frac{8}{20}+\frac{60}{20}\)
\(=\frac{77}{20}=3\frac{17}{20}\)
b) \(\left(\frac{2}{3}-\frac{3}{8}\right):\left(\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{24}:\frac{17}{20}\)
\(=\frac{35}{102}\)
c) \(\frac{15}{8}\cdot\left(\frac{1}{3}+\frac{1}{8}\right)-\frac{3}{8}:\frac{3}{4}\)
\(=\frac{15}{8}\cdot\frac{11}{24}-\frac{1}{2}\)
\(=\frac{55}{64}-\frac{1}{2}\)
\(=\frac{23}{64}\)
d) \(\frac{20}{21}:\left(\frac{4}{5}-\frac{1}{10}\right)+\frac{13}{15}:\frac{5}{26}\)
\(=\frac{20}{21}:\frac{7}{10}+\frac{52}{15}\)
\(=\frac{200}{147}+\frac{52}{15}\)
\(=4\frac{608}{735}\)
a: Sửa đề: \(\dfrac{-2}{5}\cdot\dfrac{3}{4}+\dfrac{-2}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-2}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=-\dfrac{2}{5}\cdot\dfrac{4}{4}=-\dfrac{2}{5}\)
b: \(\left(\dfrac{5}{-3}-\dfrac{1}{2}\right)\cdot\dfrac{-6}{26}\)
\(=\left(\dfrac{-5}{3}-\dfrac{1}{2}\right)\cdot\dfrac{-3}{13}\)
\(=\dfrac{-13}{6}\cdot\dfrac{-3}{13}=\dfrac{3}{6}=\dfrac{1}{2}\)
c: \(\dfrac{-4}{15}:\left[-\dfrac{2}{3}+\dfrac{3}{5}\right]\)
\(=\dfrac{-4}{15}:\dfrac{-2\cdot5+3\cdot3}{15}\)
\(=-\dfrac{4}{15}\cdot\dfrac{15}{-1}=4\)
\(-\dfrac{2}{5}\cdot\dfrac{3}{4}+\dfrac{-2}{5}+\dfrac{1}{4}=-\dfrac{2}{5}\cdot\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{2}{5}\cdot\dfrac{4}{4}=-\dfrac{2}{5}\cdot1=-\dfrac{2}{5}\)
\(\left(\dfrac{5}{-3}-\dfrac{1}{2}\right)\cdot\left(-\dfrac{6}{26}\right)=\left(-\dfrac{10}{6}-\dfrac{3}{6}\right)\cdot\left(-\dfrac{6}{26}\right)=-\dfrac{13}{6}\cdot\left(-\dfrac{6}{26}\right)=\dfrac{1}{2}\)
\(-\dfrac{4}{15}:\left[\left(-\dfrac{2}{3}\right)+\dfrac{3}{5}\right]=-\dfrac{4}{15}:\left(-\dfrac{10}{15}+\dfrac{9}{15}\right)=-\dfrac{4}{15}:\left(-\dfrac{1}{15}\right)=-\dfrac{4}{15}\cdot\left(-15\right)=4\)