Question

Tính giá trị của biểu thức:

Q = \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{3}+\sqrt{4}}{1+\sqrt{3}+\sqrt{4}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2012}+\sqrt{2013}}{1+\sqrt{2012}+\sqrt{2013}}\)

 

Answer 1

!@#$%^&*()_+\ [];'{}

đầu hàng tại chỗ !

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Answer 2

NX \(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}\)  =\(\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}-1\right)}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}+1\right)^2}\)

                                           =\(\frac{\left(\left(\sqrt{n+1}-\sqrt{n}\right)^2-1^2\right)}{n+1-n-1-2\sqrt{n}}\) \(=\frac{n+1+n-2\sqrt{\left(n+1\right)n}-1}{-2\sqrt{n}}=\frac{2n-2\sqrt{n\left(n+1\right)}}{-2\sqrt{n}}\) 

=\(\frac{n-\sqrt{n\left(n+1\right)}}{-\sqrt{n}}=\frac{n}{-\sqrt{n}}+\frac{\sqrt{n\left(n+1\right)}}{\sqrt{n}}=-\sqrt{n}+\sqrt{n+1}\)

thay vao Q ta co

Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)