Question

Cho \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)

Tính P=\(\left(a^{2017}+b^{2017}\right)\left(b^{2018}-c^{2018}\right)\)

Answer 1

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+abc+ac^2+bc^2-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+c^2\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Rightarrow\left[{}\begin{matrix}a+b=0\\a+c=0\\b+c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=-b\\c=-a\\b=-c\end{matrix}\right.\)TH1: nếu a=-b

P=(a²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=(-b²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=0

TH2: nếu b=-c

P=(a²⁰¹⁷+b²⁰¹⁷)(b²⁰¹⁸-c²⁰¹⁸)=(a²⁰¹⁷+b²⁰¹⁷)((-c)²⁰¹⁸-c²⁰¹⁸)=0

Còn một TH nữa thì bạn ghi thiếu đề rồi