Question

Tìm các số a,b,c biết : \(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{1}{a+b+c}\)

Answer 1

áp dụng t.c dãy tỉ số bằng nhau ta có:

\(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{b+c+1+a+c+2+a+b-3}{a+b+c}=2\)(vì a+b+c khác 0)

\(\Rightarrow\frac{1}{a+b+c}=2\Rightarrow a+b+c=\frac{1}{2}\)

\(\frac{b+c+1}{a}=2\Rightarrow2a=b+c+1\Rightarrow3a=a+b+c+1\Rightarrow a=\frac{1}{2}\)

\(\frac{a+c+2}{b}=2\Rightarrow2b=a+c+2\Rightarrow3b=a+b+c+2\Rightarrow b=\frac{5}{6}\)

\(\frac{a+b-3}{c}=2\Rightarrow2c=a+b-3\Rightarrow3c=a+b+c-3\Rightarrow c=-\frac{5}{6}\)

Vậy \(a=\frac{1}{2},b=\frac{5}{6},c=-\frac{5}{6}\)