Question

[Vịtt Tên Hiền]

cho a³+b³+c³=3abc với\(a,b,c\ne0\)

tính giá trị của biểu thức P=\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

Answer 1

[Hoang Hung Quan]

Giải:

Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Ta xét các trường hợp:

Trường hợp \(1\): Nếu \(a+b+c=0\) thì:

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

Thay vào \(P\) ta có:

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

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

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)

Trường hợp \(2\): Nếu \(a=b=c\) thì:

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

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

\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)

\(=2.2.2=8\)

Vậy \(P=-1\) hoặc \(P=8\)

Answer 2

[Trần Ánh Ngọc]

ta có : a³+b³+c³-3abc=0

\(\Rightarrow\)(a+b)³+c³-3abc-3a²b-3ab²=0

\(\Rightarrow\)(a+b+c)(a²+b²+c²+2ab-ac-bc)-3ab(a+b+c)=0

\(\Rightarrow\)(a+b+c)(a²+b²+c²-ab-ac-bc)=0

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

\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\\\left(a+b+c\right)^2+a^2+b^2+c^2=0\Leftrightarrow a=b=c=0\left(bỏ\right)\end{matrix}\right.\)ta có P=(1+\(\dfrac{a}{b}\))(1+\(\dfrac{b}{c}\))(1+\(\dfrac{c}{a}\))

\(\Leftrightarrow\)p=\(\left(\dfrac{b+a}{b}\right)\left(\dfrac{c+b}{c}\right)\left(\dfrac{a+c}{a}\right)\)

\(\Leftrightarrow P=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)\)

\(\Leftrightarrow\)P=-1