Đặt \(2017=a\)

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

Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)

\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)

Ta có : 3 - |x - 2017| - (x - 2017)² 

= 3 - [|x - 2017| + (x - 2017)² ]

Mà \(\left|x-2017\right|\ge0\forall x\)

      \(\left(x-2017\right)^2\ge0\forall x\)

=>  [|x - 2017| + (x - 2017)² ] \(\ge0\forall x\)

Nên :  3 - [|x - 2017| + (x - 2017)² ] hay  3 - |x - 2017| - (x - 2017)²  \(\le3\forall x\)

Vậy GTLN của biểu thức là 3 khi và chỉ khi x = 2017

-A = (x-2017)^2 + /x-2017/ - 3

= (/x-2017/+1/2)^2 - 7/2 

>= -7/2

=> A <= 7/2

Dấu "=" xảy ra <=> x-2017 = 0 <=> x= 2017 

Vậy Max A= 7/2 <=> x= 2017

cm \(a^3+b^3+c^3=3abc\)

thì \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

(chuyển vế xét hiệu ) 

TA CÓ: \(a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow a-b=0;c-a=0;b-c=0\Rightarrow a=b=c\)

\(\Rightarrow\frac{a^{2017}}{b^{2017}}+\frac{b^{2017}}{c^{2017}}+\frac{c^{2017}}{a^{2017}}=1+1+1=3\)

Lời giải:

Tại $x=2016$ thì $x-2016=0$

Khi đó:
$A=x^{2016}(x-2016)-x^{2015}(x-2016)+x^{2014}(x-2016)-x^{2013}(x-2016)+.....-x(x-2016)+x-2017$

$=x^{2016}.0-x^{2015}.0+......-x.0+2016-2017=2016-2017=-1$

\(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow a=b=c\) (a,b,c là các số dương)

Bạn thay vào A để tính.

Ta có:

a²⁰¹⁷ + b²⁰¹⁷ = a²⁰¹⁷ + ab²⁰¹⁶ + a²⁰¹⁶b + b²⁰¹⁷ - a²⁰¹⁶b - ab²⁰¹⁶

= a.(a²⁰¹⁶ + b²⁰¹⁶) + b.(b²⁰¹⁶ + a²⁰¹⁶) - ab.(a²⁰¹⁵ - b²⁰¹⁵)

= (a²⁰¹⁶ + b²⁰¹⁶).(a + b) - ab.(a²⁰¹⁵ + b²⁰¹⁵)

Chia cả 2 vế cho a²⁰¹⁷ + b²⁰¹⁷ = a²⁰¹⁶ + b²⁰¹⁶ = a²⁰¹⁵ + b²⁰¹⁵

=>  a + b - ab = 1

=> a.(1 - b) - 1 + b  = 0

=> a.(1 - b) - (1 - b) = 0

=> (1 - b).(a - 1) = 0

=> a = b = 1

Ta có: P = 20.a + 11.b + 2017

P = 20.1 + 11.b + 2017

P = 20 + 11 + 2017

P = 2048