\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)

\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)

=> Bằng nhau

\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)

\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)

vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)

\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

2017>2015

=>căn 2017>căn 2015

=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)

=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)

Xét với x > 0 : \(\sqrt{1+\left(x-1\right)^2+\frac{\left(x-1\right)^2}{x^2}}+\frac{x-1}{x}=\sqrt{\frac{\left(x^2-x+1\right)^2}{x^2}}+\frac{x-1}{x}\)

\(=\frac{x^2-x+1}{x}+\frac{x-1}{x}=\frac{x^2}{x}=x\)

Áp dụng với x = 2017 suy ra biểu thức cần tính có giá trị bằng 2017

\(A=\sqrt{2016^2+\frac{2017}{2017}+\frac{2016^2-1}{2017^2}-\frac{1}{2017^2}}+\frac{2016}{2017}\)

\(A=\sqrt{2016^2+\frac{1}{2017^2}+\frac{2015.2017}{2017^2}+\frac{2017}{2017}}+\frac{2016}{2017}\)

\(A=\sqrt{2016^2+2.2016.\frac{1}{2017}+\frac{1^2}{2017^2}}+\frac{2016}{2017}\)

\(A=\sqrt{\left(2016+\frac{1}{2017}\right)^2}+\frac{2016}{2017}\)

\(A=\left(2016+\frac{1}{2017}\right)+\frac{2016}{2017}\)

A = 2017

Chúc bạn làm bài tốt

Với mọi \(n\in N.\)ta có:

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

\(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}.=1-\frac{1}{\sqrt{2017}}=\frac{\sqrt{2017}-1}{\sqrt{2017}}.\)

a )\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)

=\(\sqrt{2+3+1+2\sqrt{2.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}}\)

=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}\)

=\(\sqrt{2}+\sqrt{3}+1\)

\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)