Violympic toán 9

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Khuất Tuấn Hùng

\(\left\{{}\begin{matrix}ab+a+b=3\\bc+b+c=8\\ca+c+a=15\end{matrix}\right.\)

Hà Nam Phan Đình
2 tháng 12 2017 lúc 20:25

HPT tương đương

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)\left(b+1\right)=4\left(1\right)\\\left(b+1\right)\left(c+1\right)=9\left(2\right)\\\left(a+1\right)\left(c+1\right)=16\left(3\right)\end{matrix}\right.\)

lấy (1).(2).(3)

\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2=576\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=24\left(4\right)\)

Lấy \(\dfrac{\left(4\right)}{\left(1\right)};\dfrac{\left(4\right)}{\left(2\right)};\dfrac{\left(4\right)}{\left(3\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}c+1=6\\a+1=\dfrac{8}{3}\\b+1=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3}\\b=\dfrac{1}{2}\\c=5\end{matrix}\right.\)