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\(\frac{-7}{21} +\left(1+\frac{1}{3}\right)\)
\(=\frac{-1}{3}+1+\frac{1}{3}=1\)
5.7.77-7.60+49.25-15.42
= 5.7.77-7.5.12+7.7.5.5-3.5-3.5.6.7
= 35.77-35.12+35.35-35.18
= 35.(77-12+35-18)
= 35.82
= 2870
2.
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2017}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ \dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}=\dfrac{2016}{2017}:2\\ \dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}=\dfrac{1008}{2017}\\ \dfrac{1}{\left(x+1\right)}=\dfrac{1}{2}-\dfrac{1008}{2017}\\ \dfrac{1}{\left(x+1\right)}=\dfrac{1}{4034}\\ \Rightarrow x+1=4034\Rightarrow x=4033\)