Question

Thực hiện phép tính

a)\(\sqrt{1+\left(\dfrac{1}{2017}+\dfrac{1}{2019}\right)^2}\)

b)\(\sqrt{2017^2+2017^2.2018^2+2018^2}\)

Answer 1

a: Đặt a=2017

\(A=\sqrt{1+\left(\dfrac{1}{a}+\dfrac{1}{a+2}\right)^2}\)

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

\(=\sqrt{1+\dfrac{4a^2+8a+4}{a^2\cdot\left(a+2\right)^2}}=\sqrt{\dfrac{\left(a^2+a\right)^2+4a^2+8a+4}{a^2\left(a+2\right)^2}}\)

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

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

\(=\dfrac{\sqrt{\left(2017^2+2017\right)^2+4\cdot2018^2}}{2017\cdot2019}\)

b: Đặt 2017=a

\(B=\sqrt{a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2}\)

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

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

\(=2017^2+2017+1=4070307\)