Question

cmr B là số hữu tỉ \(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

Answer 1

\(\sqrt{1+a^2+\left(\frac{a}{a+1}\right)^2}\)=\(\sqrt{\frac{\left(a+1\right)^2+a^2.\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}\) =\(\sqrt{\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}\)

=\(\sqrt{\frac{a^4+2a^2.\left(a+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}\) =\(\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}=\frac{a^2+a+1}{a+1}=\frac{a\left(a+1\right)+1}{a+1}=a+\frac{1}{a+1}\)

thay vao dau bai ta co 

\(2017+\frac{1}{2018}+\frac{2017}{2018}=2017+1=2018\)

Answer 2

khó quá nhỉ