Question

[David Santas]

Cho a, b,c thỏa mãn: \(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}\frac{c+4a-b}{b}\)

Tính P = \(\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)

Answer 1

[Diệu Huyền]

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

\(\frac{1}{a+b+c}=\frac{a+4b-c+b+4c-a+c+4a-b}{a+b+c}\)

\(=\frac{4\left(a+b+c\right)}{a+b+c}=4\)

\(\Rightarrow\left\{{}\begin{matrix}4c=a+4b-c\\4a=b+4a-a\\4b=c+4a-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5c=a+4b\\5a=b+4c\\5b=c+4a\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

\(P=\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)

\(=\left(2+1\right)\left(3+1\right)\left(4+1\right)\)

\(=3.4.5=60\)

Vậy .............

Cái đề thiếu dấu " = " kìa -__-

Answer 2

[David Santas]

hì