\(x^2-2\left(2m+1\right)+3m=0\)
\(B=\dfrac{S}{P}=\dfrac{2\left(2m+1\right)}{3m}=\dfrac{4m+2}{3m}\in Z\)
\(\Leftrightarrow4m+2⋮3m\)
\(\Leftrightarrow3\left(4m+2\right)-4.3m⋮3m\)
\(\Leftrightarrow6⋮3m\)
\(\Leftrightarrow2⋮m\) \(\left(m\in Z\right)\)
\(\Leftrightarrow m\in\left\{\pm1;\pm2\right\}\) thỏa mãn đề bài
\(\text{Δ}=\left[2\left(2m+1\right)\right]^2-4\cdot1\cdot3m\)
\(=\left(4m+2\right)^2-12m=16m^2+16m+4-12m\)
\(=16m^2+4m+4=4\left(4m^2+m+1\right)\)
\(=4\left(4m^2+2\cdot2m\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right)\)
\(=4\left(2m+\dfrac{1}{4}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(2m+1\right)=4m+2\\x_1x_2=\dfrac{c}{a}=3m\end{matrix}\right.\)
Để B là số nguyên thì \(x_1+x_2⋮x_1x_2\)
=>\(4m+2⋮3m\)
=>\(12m+6⋮3m\)
=>\(6⋮3m\)
mà m nguyên
nên \(3m\in\left\{3;-3;6;-6\right\}\)
=>\(m\in\left\{1;-1;2;-2\right\}\)
Thay lại vào trong B, ta được: \(m\in\left\{1;-2\right\}\)
