Question

Tìm 3 số a,b,c biết : \(\dfrac{a-1}{2}=\dfrac{b-2}{3}=\dfrac{c-3}{4}\) và a - 2b + 3c = 14

Answer 1

Vì \(\dfrac{a-1}{2}=\dfrac{b-2}{3}=\dfrac{c-3}{4}\)

nên \(\dfrac{a-1}{2}=\dfrac{2b-4}{6}=\dfrac{3c-9}{12}\)

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

\(\dfrac{a-1}{2}=\dfrac{2b-4}{6}=\dfrac{3c-9}{12}=\dfrac{a-1-2b+4+3c-9}{2-6+12}=\dfrac{14-6}{8}=1\)

Do \(\dfrac{a-1}{2}=1\Rightarrow a=3\)

\(\dfrac{2b-4}{6}=1\Rightarrow b=5\)

\(\dfrac{3c-9}{12}=1\Rightarrow c=7\)

Vậy \(\left\{{}\begin{matrix}a=3\\b=5\\c=7\end{matrix}\right..\)