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

http://olm.vn/hoi-dap/question/305454.html

sao lại thế

khó nhỉ!!!

gggggggggggggggggggggg

theo bài ra ta có \(a^{2012}+b^{2012}=a^{2013}+b^{2013}=a^{2014}+b^{2014}\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\)\(\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\Leftrightarrow\)

\($\left(a^{1006}-a^{1007}\right)^2+\left(b^{1006}-b^{1007}\right)=0$\)

\(\Leftrightarrow\left\{\begin{matrix}a^{1006}-a^{1007}=0\\b^{1006}-b^{1007}=0\end{matrix}\right.\left\{\begin{matrix}a=0;a=1\\b=0;b=1\end{matrix}\right.\)

Khi đó P=20.0+11.0+2013=2013

hoặc P=20.1+11.0+2013=2033

hoặc p=20.0+11.1+2013=2024

\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_{2013}}{a_{2014}}=\dfrac{a_{2014}}{a_1}=\dfrac{a_1+a_2+...+a_{2014}}{a_1+a_2+...+a_{2014}}=1\\ \Leftrightarrow a_1=a_2=...=a_{2014}\\ \Leftrightarrow Q=\dfrac{\left(2014a_1\right)^2}{a_1^2\left(1+2+...+2014\right)}=\dfrac{2014^2\cdot a_1^2}{a_1^2\cdot\dfrac{2015\cdot2014}{2}}=\dfrac{2\cdot2014^2}{2015\cdot2014}=\dfrac{2\cdot2014}{2015}=...\)