Question

Cho a,b,c thỏa mãn \(\dfrac{1}{2a-1}\)=\(\dfrac{2}{3b-1}\)=\(\dfrac{3}{4c-1}\) và 3a+2b-c =4.Tìm a,b,c

Answer 1

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

\(\Rightarrow\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}\)và 3a+2b-c=4

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}=\dfrac{36a-18+24b-8-12c+3}{18+16-9}=\dfrac{12\left(3a+2b-c\right)-23}{25}=\dfrac{12\cdot4-23}{25}=1\)

=>2a-1=1<=>a=1

3b-1=2<=>b=1

4c-1=3<=>c=1

Vậy...

Answer 2

\(\dfrac{1}{2a-1}=\dfrac{2}{3b-1}=\dfrac{3}{4c-1}\\ \Rightarrow\dfrac{2a-1}{1}=\dfrac{3b-1}{2}=\dfrac{4c-1}{3}\\ \Rightarrow\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}=\dfrac{36a-18+24b-8-12c+3}{18+16-9}=\dfrac{12\left(3a+2b-c\right)-23}{25}\\ =\dfrac{12.4-23}{25}=1\\ \Rightarrow a=1;b=1;c=1\)