Question

Cho a,b,c là ba số thảo mãn: abc=1 và \(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Tính giá trị biểu thức :

P=\(\left(a^{2019}-1\right)\left(b^{2020}-1\right)\left(c^{2021}-1\right)\)

Answer 1

Lời giải:

Do $abc=1$ nên:

$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=bc+ac+ab$

$\Leftrightarrow ab+bc+ac-a-b-c=0$

$\Leftrightarrow (ab-a-b+1)+bc+ac-c-1=0$

$\Leftrightarrow (ab-a-b+1)+bc+ac-c-abc=0$

$\Leftrightarrow (ab-a-b+1)+c(b+a-1-ab)=0$

$\Leftrightarrow (ab-a-b+1)(1-c)=0$

$\Leftrightarrow (a-1)(b-1)(1-c)=0$

$\Leftrightarrow (a-1)(b-1)(c-1)=0$

Do đó:

$P=(a^{2019}-1)(b^{2019}-1)(c^{2019}-1)=(a-1)(a^{2018}+...+1)(b-1)(b^{2019}+...+1)(c-1)(c^{2020}+...+1)$

$=(a-1)(b-1)(c-1).(a^{2018}+...+1)(b^{2019}+...+1)(c^{2020}+...+1)=0$