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a: =17-43=-26
b: =-34+34+11-11+105=105
c: \(=64+125\cdot100=12564\)
d: \(=60+\left[7^3-343\right]\cdot2017^{2018}=60\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{\frac{2017}{2019}}{2}=\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}=\frac{1}{2019}\)
=> x + 1 = 2019 <=> x = 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~