Question

tính \(A=3\sqrt{5}-\sqrt{\dfrac{1}{5}}+\dfrac{3}{\sqrt{5}-1}\)

\(B=\sqrt{2018^2+2018^22019^2+2019^2}\)

Answer 1

\(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3\left(\sqrt{5}+1\right)}{5-1}\)

\(=\dfrac{14}{5}\sqrt{5}+\dfrac{3}{4}\sqrt{5}+\dfrac{3}{4}\)

\(=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)

Answer 2

a: \(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3}{4}+\dfrac{3}{4}\sqrt{5}=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)

b: Đặt a=2018

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

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

\(=\sqrt{2a^2+1+2a+a^4+2a^3+a^2}\)

\(=\sqrt{a^4+2a^3+3a^2+2a+1}\)

\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1=2018^2+2018+1\)