Question

[binh2k5]

cho 2 bieu thuc:

\(A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\) va \(B=\frac{x+1}{x+2}\)

a.Rut gon A

b.tim gia tri nguyen cua x de P=A:B co gia tri nguyen

Answer 1

[❤ Hoa ❤]

\(A=\frac{3}{2-x}+\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)

\(A=\frac{-3}{x-2}+\frac{3}{x+2}+\frac{3x^2}{\left(x+2\right)\left(x-2\right)}\)

\(A=\frac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{-3x-6+3x-6+3x^2}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{-12+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(-4+x^2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(A=3\)

Answer 2

[Incursion_03]

\(a,A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)

       \(=\frac{-3\left(x+2\right)-3\left(x-2\right)+3x^2}{\left(x-2\right)\left(x+2\right)}\)

       \(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)

       \(=\frac{3x^2-6x}{\left(x-2\right)\left(x+2\right)}\)

      \(=\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

      \(=\frac{3x}{x+2}\)

\(b,ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm2\\x\ne-1\end{cases}}}\)

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 \(x\in\left\{-4;-2;0;2\right\}\)