\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)

hình như cái này đâu phải toán lớp 5 đâu bạn

nhầm toán lớp 6

12+13×14

giải đc chết liền

\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)

\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)

\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)

\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

\(M=1^2-2^2+3^2-4^2+...-2016^2+2017^2\)

\(=\left(2017^2-2016^2\right)+...+\left(3^2-2^2\right)+1^2\)

\(=\left(2017-2016\right)\left(2017+2016\right)+...+\left(3-2\right)\left(3+2\right)+1\)

\(=2017+2016+...+3+2+1\)

\(=\frac{2017\cdot\left(2017+1\right)}{2}=2035153\)

99/100 đó

\(M=1^2-2^2+3^2-4^2+...-2016^2+2017^2\)

\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(2015^2-2016^2\right)+2017^2\)

\(=\left(1+2\right)\left(1-2\right)+\left(3+4\right)\left(3-4\right)+...+\left(2015+2016\right)\left(2015-2016\right)+2017^2\)

\(=\left(1+2\right)\cdot\left(-1\right)+\left(3+4\right)\cdot\left(-1\right)+...+\left(2015+2016\right)\cdot\left(-1\right)+2017^2\)

\(=\left(-1\right)\cdot\left(1+2+...+2016\right)+2017^2\)

\(=\left(-1\right)\cdot\frac{2016\cdot\left(2016+1\right)}{2}+2017^2\)

\(=-2033136+4068289=2035153\)

\(M=\frac{2017.2018}{2}\)

Tầm nhìn quá xa

2035153