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Để P là số nguyên thì \(3x^3-5x^2+9x-15-1⋮3x-5\)
\(\Rightarrow3x-5\in\left\{1;-1\right\}\)
=>x=2(vì x là số nguyên)
a) Sửa đề
\(A=\dfrac{2x^2-8}{x^2-4x+4}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)^2}=\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)^2}\)
\(A=\dfrac{2\left(x+2\right)}{x-2}\)
b) Ta có :
A = \(\dfrac{2x+4}{x-2}=\dfrac{2x-4+8}{x-2}=\dfrac{2\left(x-2\right)+8}{x-2}=\dfrac{2\left(x-2\right)}{x-2}+\dfrac{8}{x-2}\)
A = \(2+\dfrac{8}{x-2}\)
Để A có giá trị nguyên thì :
x - 2 thuộc Ư(8)
Từ đó , ta có bảng giá trị sau :
n-2 n 1 -1 2 -2 4 -4 8 -8 3 1 4 0 6 -2 10 -6
Vậy ,....
\(A=\frac{3}{2-x}+\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(A=\frac{-3}{x-2}+\frac{3}{x+2}+\frac{3x^2}{\left(x+2\right)\left(x-2\right)}\)
\(A=\frac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-3x-6+3x-6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-12+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(-4+x^2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(A=3\)
\(a,A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(=\frac{-3\left(x+2\right)-3\left(x-2\right)+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x^2-6x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x}{x+2}\)
\(b,ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm2\\x\ne-1\end{cases}}}\)
Ta có : \(P=A:B=\frac{3x}{x+2}:\frac{x+1}{x+2}\)
\(=\frac{3x}{x+2}.\frac{x+2}{x+1}\)
\(=\frac{3x}{x+1}\)
\(=\frac{3x+3}{x+1}-\frac{3}{x+1}\)
\(=3-\frac{3}{x+1}\)
Để P nguyên thì \(3-\frac{3}{x+1}\inℤ\)
\(\Leftrightarrow\frac{3}{x+1}\inℤ\)
Vì \(x\inℤ\Rightarrow x+1\inℤ\)
Ta có bảng :
| x + 1 | -3 | -1 | 1 | 3 |
| x | -4 | -2 | 0 | 2 |
Vậy \(x\in\left\{-4;-2;0;2\right\}\)