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a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
\(\dfrac{x-7}{7}=\dfrac{y-6}{6}=\dfrac{x-y-1}{1}=\dfrac{-4-1}{1}=-5\)
\(\Rightarrow x-7=7.\left(-5\right)=-35\Rightarrow x=-28;y-6=6.\left(-5\right)=-30\Rightarrow y=-24\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(\Leftrightarrow6x-42=7y-42\)
\(\Leftrightarrow6x=7y\)
\(\Leftrightarrow6\left(x-y\right)-y=0\)
\(\Leftrightarrow-24-y=0\)
\(\Leftrightarrow y=-24\)
\(\Leftrightarrow\dfrac{x-7}{-24-6}=\dfrac{7}{6}\Leftrightarrow\dfrac{x-7}{-30}=\dfrac{7}{6}\Leftrightarrow6x-30=-210\Leftrightarrow6x=-180\Leftrightarrow x=-30\)
\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2009}{2010}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2009}{2010}\)
\(\Leftrightarrow-\dfrac{1}{x+1}=-\dfrac{1}{2010}\)
\(\Leftrightarrow x+1=2010\Leftrightarrow x=2009\)
Bài 1
a) 25 - 5x = 5
=> - 5x = 5 - 25
=> - 5x = -20
=> x = 4