Question

[cherry moon]

\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\)\(\frac{1}{\sqrt{2005}+\sqrt{2009}}\)

Rút gọn biểu thức A

\(B=x^3-3x+2000\). Rút gọn B biết \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

Mong mọi người giúp đỡ mình ạ , mình rất cần ạ

Answer 1

[Nguyễn Linh Chi]

1. Trục căn thức ở mẫu:

\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)

=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)

\(=\frac{\sqrt{2009}-1}{4}\)

2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)

\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)

\(=6+3x\)

=> \(x^3-3x=6\)

=> \(B=x^3-3x+2000=6+2000=2006\)

Answer 2

[Nguyễn Ngọc Anh Minh]

\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)

\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)

\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)