cho a,b,c thuộc r khác 0. thỏa mãn đk: a+b-c/c=b+c-a/a=c+a-b/b. tính gtbt:p=(1+b/a)(1+a/c)(1+c/b)

Đáp án:

\(P = {\left( {\frac{{1 + a}}{a}} \right)^3}\)

Giải thích các bước giải:

\(\begin{array}{l}
\frac{{a + b – c}}{c} = \frac{{b + c – a}}{a} = \frac{{c + a – b}}{b} = \frac{{a + b – c + b + c – a + c + a – b}}{{a + b + c}} = 1\\
 \to \left\{ \begin{array}{l}
a + b – c = c\\
b + c – a = a\\
c + a – b = b
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 2c – b\\
b + c – 2a = 0\\
c + a – 2b = 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 2c – b\\
b + c – 2(2c – b) = 0\\
c + 2c – b – 2b = 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 2c – b\\
3b – 3c = 0\\
3c – 3b = 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 2c – c = c\\
b = c
\end{array} \right. \leftrightarrow a = b = c\\
P = \frac{{1 + b}}{a}.\frac{{1 + a}}{c}.\frac{{1 + c}}{b} = \frac{{1 + a}}{a}.\frac{{1 + a}}{a}.\frac{{1 + a}}{a} = {\left( {\frac{{1 + a}}{a}} \right)^3}
\end{array}\)