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

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

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

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

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

A = 2017

Chúc bạn làm bài tốt

Đặt 2017 = a thì ta có 

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

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

= a

Vậy cái đó bằng 2017

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

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

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

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

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

Đặt C = 1 + 2017 + 2017² + ... + 2017²⁰¹⁶ ; D = 1 + 2016 + 2016² + ... + 2016²⁰¹⁶

Ta có : 2017C = 2017 + 2017² + 2017³ + ... + 2017²⁰¹⁷

=> 2016C = 2017C - C = 2017²⁰¹⁷ - 1\(\Rightarrow C=\frac{2017^{2017}-1}{2016}\)

2016D = 2016 + 2016² + 2016³ + ... + 2016²⁰¹⁷

=> 2015D = 2016D - D = 2016²⁰¹⁷ - 1\(\Rightarrow D=\frac{2016^{2017}-1}{2015}\)

\(\Rightarrow A=\frac{2017^{2017}}{\frac{2017^{2017}-1}{2016}}=\frac{2017^{2017}.2016}{2017^{2017}-1}\);\(B=\frac{2016^{2017}}{\frac{2016^{2017}-1}{2015}}=\frac{2016^{2017}.2015}{2016^{2017}-1}\)

Ta có : 2017²⁰¹⁷.2016.(2016²⁰¹⁷ - 1) - 2016²⁰¹⁷.2015.(2017²⁰¹⁷ - 1)

= 2017²⁰¹⁷.2016²⁰¹⁷.2016 - 2017²⁰¹⁷.2016 - 2017²⁰¹⁷.2016²⁰¹⁷.2015 + 2016²⁰¹⁷.2015

= 2017²⁰¹⁷.2016²⁰¹⁷ - 2017²⁰¹⁷.2016 + 2016²⁰¹⁷.2015

= 2017²⁰¹⁷.(2016²⁰¹⁷ - 2016) + 2016²⁰¹⁷.2015 > 0

=> A > B

Ta có 

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

\(B=1:\frac{1+2016+2016^2+...2016^{2016}}{2016^{2017}}\)

\(A=1:\left(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}\right)\)

\(B=1:\left(\frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\right)\)

Có 2017²⁰¹⁷>2016²⁰¹⁷ ;  2017²⁰¹⁶>2016²⁰¹⁶ ;  2017²⁰¹⁵>2016²⁰¹⁵;..... ; 2017>2016

=> \(\frac{1}{2017^{2017}}< \frac{1}{2016^{2017}};\frac{1}{2017^{2016}}< \frac{1}{2016^{2016}};\frac{1}{2017^{2015}}< \frac{1}{2016^{2015}};...;\frac{1}{2017}< \frac{1}{2016}\)

=> \(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}< \frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\)

=> A>B ( vì số bị chia và số chia của A và B đều dương, số bị chia của cả 2 đều là 1, cái nào có số chia nhỏ hơn thì lớn hơn)

Xét biểu thức  \(N=1+k+k^2+k^3+...+k^n\) (1) với k là số tự nhiên lớn hơn 1

Ta có \(k.N=k+k^2+k^3+k^4+...+k^{n+1}\) (2)

Lấy (2) - (1) ta được:

\(\left(k-1\right)N=\left(k+k^2+k^3+k^4+...+k^{n+1}\right)-\left(1+k+k^2+k^3+...+k^n\right)=k^{n+1}-1\)

Suy ra  \(N=\frac{k^{n+1}-1}{k-1}\) 

Áp dụng với k = 2017; n = 2016 ta được \(1+2017+2017^2+...+2017^{2016}=\frac{2017^{2017}-1}{2016}\)

Áp dụng với k = 2016; n = 2016 ta được \(1+2016+2016^2+...+2016^{2016}=\frac{2016^{2017}-1}{2015}\)

\(A=\frac{2017^{2017}}{1+2017+2017^2+...+2017^{2016}}=\frac{2017^{2017}}{\frac{2017^{2017}-1}{2016}}=\frac{2016.2017^{2017}}{2017^{2017}-1}>1\) 

Tương tự  \(B=\frac{2015.2016^{2017}}{2016^{2017}-1}>1\)

Mặt khác: Tử số A > tử số B; mẫu A > mẫu B => A < B.

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

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

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

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

Ta có:

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

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

Suy ra:

\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)

Vậy Q < R.