Question

\(\frac{a}{b}\)=\(\frac{b}{c}\)= \(\frac{c}{a}\)biết a+b+c khác 0

tính M=\(\frac{a^{2020}.b^2.c}{c^{2023}}\)giúp mình với

 

Answer 1

Ta đặt

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\left(k\in R\right)\)

=>a=bk;b=ck;c=ak

=>a+b+c=k(a+b+c) 

Mà a+b+c khác 0

=>1=k

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)

=>a=b=c

=>M=\(\frac{a^{2020}.b^2.c}{c^{2023}}=\frac{a^{2020}.a^2.a}{a^{2023}}=\frac{a^{2023}}{a^{2023}}=1\)

Vậy M=1

Answer 2

tu day bieu thu => a=b=c

M=a^(2020+2+1)/a^2023=a^2023/a^2023

M=1