\(\frac{3x^3+13x^2-7x+5}{3x-2}=\frac{x^2\left(3x-2\right)+5x\left(3x-2\right)+\left(3x-2\right)+7}{3x-2}=x^2+5x+1+\frac{7}{3x-2}\)
Để p/s trên nhận giá trị nguyên thì 3x-2 thuộc ước của 7
Bạn tự liệt kê
\(\frac{3x^3+13x^2-7x+5}{3x-2}=\frac{x^2\left(3x-2\right)+15x^2-7x+5}{3x-2}=\frac{x^2\left(3x-2\right)}{3x-2}+\frac{15x^2-7x+5}{3x-2}=x^2+\frac{15x^2-7x+5}{3x-2}\in Z\)
\(\Rightarrow\frac{15x^2-7x+5}{3x-2}=\frac{5x\left(3x-2\right)+3x+5}{3x-2}=\frac{5x\left(3x-2\right)}{3x-2}+\frac{3x+5}{3x-2}=5x+\frac{3x+5}{3x-2}\in Z\)
\(\Rightarrow\frac{3x+5}{3x-2}=\frac{3x-2+7}{3x-2}=\frac{3x-2}{3x-2}+\frac{7}{3x-2}=1+\frac{7}{3x-2}\in Z\)
\(\Rightarrow7⋮3x-2\)
\(\Rightarrow3x-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{1;3\right\}\) vì x thuộc Z
Ta có :
Do đó để \(\frac{3x^3+13x^2-7x+5}{3x-2}\in Z\Rightarrow3x-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{-\frac{5}{3};-\frac{1}{3};1;3\right\}\)
Mà \(x\in Z\Rightarrow x\in\left\{1;3\right\}\)
