\(P=\sqrt{1+999^2+\dfrac{999^2}{1000^2}+\dfrac{999}{1000}}\)

\(\Leftrightarrow\)\(\sqrt{\dfrac{1999}{1000}+999^2+\dfrac{999^2}{1000^2}}\)

???

=4,576491223

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

 =\(16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}\) =\(16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(\Rightarrow A=\sqrt{12+4\sqrt{5}}\)

Cm giúp mình với nhé. Mình cảm ơn ạ!

đề bài là j vậy bn

Áp dụng \(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\) ta có:

\(x=\sqrt{1+\dfrac{1}{\left(\dfrac{1}{999}\right)^2}+\dfrac{1}{\left(\dfrac{1}{999}+1\right)^2}}+\dfrac{999}{1000}=1+\dfrac{1}{\dfrac{1}{999}}-\dfrac{1}{\dfrac{1}{999}+1}+\dfrac{999}{1000}=1+999-\dfrac{999}{1000}+\dfrac{999}{1000}=1000\)

???

Đề bài khó quá làm sao đây

Đặt S = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{8-2\sqrt{10-2\sqrt{5}}}\)

S² = 8 + 2\(\sqrt{10+2\sqrt{5}}\) + \(8-2\sqrt{10+2\sqrt{5}}\) + 2\(\times\)\(\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)

= 16 + \(2\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

= 16 + 2 \(\sqrt{64-40+8\sqrt{5}}\)

= 16 + 2\(\sqrt{20+2\times2\sqrt{5}\times2+4}\)

= 16 + 2\(\sqrt{\left(\sqrt{20}+2\right)^2}\)

= 16 + 2\(\sqrt{20}-4\)

= 12 + 2\(\sqrt{20}\)

Do S > 0 nên

S = \(\sqrt{12+2\sqrt{20}}\)= \(\sqrt{12+2\times2\sqrt{5}}\)=\(\sqrt{4\left(3+\sqrt{5}\right)}\)=\(2\sqrt{3+\sqrt{5}}\)

Vậy S = 2\(\sqrt{3+\sqrt{5}}\)