Ta có:
\(\left(a-\dfrac{1}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\) (1)
Và: \(a+1=b+2=c+3\)
\(\Rightarrow a=b+2-1=b+1\)
Thay vào (1) ta có:
\(\left(b+1-\dfrac{1}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\)
\(\Rightarrow\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\) (2)
Mà: \(b+2=c+3\)
\(\Rightarrow c=b+2-3=b-1\)
Thay vào (2) ta có:
\(\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(b-1-3\right)=0\)
\(\Rightarrow\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(b-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=-\dfrac{2}{3}\\b=-\dfrac{1}{2}\\b=4\end{matrix}\right.\)
TH1 khi b=\(-\dfrac{2}{3}\)
\(\Rightarrow a=b+1=-\dfrac{2}{3}+1=\dfrac{1}{3}\)
\(\Rightarrow c=b-1=-\dfrac{2}{3}-1=-\dfrac{5}{3}\)
TH2 khi \(b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+1=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
\(\Rightarrow c=b-1=-\dfrac{1}{2}-1=-\dfrac{3}{2}\)
TH3 khi \(b=4\)
\(\Rightarrow a=b+1=4+1=5\)
\(\Rightarrow c=b-1=4-1=3\)
Vậy: ...
