Question

Cho \(\dfrac{a+b-2017c}{c}=\dfrac{b+c-2017a}{a}=\dfrac{c+a-2017b}{b}\)

Với a, b, c \(\ne0\). TínhP = \(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)\)

Answer 1

Trần Thọ Đạt ông giải dùm đi!Bn ý k bk tag nên tui tag dùm!

Answer 2

Trần Thọ Đạt, giải giúp mình

Answer 3

TH1 : Nếu a + c+ b \(\ne\) 0

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{a+b-2017c}{c}=\dfrac{b+c-2017a}{a}=\dfrac{c+a-2017b}{b}=\dfrac{a+b-2017c+b+c-2017a+c+a-2017b}{c+a+b}=\dfrac{-2015a-2015b-2015c}{a+b+c}=\dfrac{-2015.\left(a+b+c\right)}{a+b+c}=-2015\)( vì a + b + c \(\ne\)0 )

\(\Rightarrow\left\{{}\begin{matrix}a+b-2017c=-2015c\\b+c-2017a=-2015a\\c+a-2017b=-2015b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)

Mặt khác , P = \(\left(\dfrac{b+a}{b}\right).\left(\dfrac{c+b}{c}\right).\left(\dfrac{a+c}{a}\right)\)

\(\Rightarrow\)\(\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=\dfrac{2.âbc}{abc}=2\)

TH2 : Nếu a + b+ c = 0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)=-1

Vây ...

Mk ko chắc là TH2 đúng ko nữa