Question

Cho biểu thức:

P= 2x^2/x^2-1 + x/x+1 - x/x-1

a, Tìm x để biểu thức P có nghĩa

b, Rút gọn P

c, Tính P tại x= -3

d, Tìm giá trị nguyên của x để P có giá trị nguyên

Answer 1

a) ĐKXĐ của P

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne-1;x\ne1\\x\ne-1\\x\ne1\end{matrix}\right.\\ \Leftrightarrow x\ne-1;x\ne1\)

Answer 2

a) P được xác định \(\Leftrightarrow\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x-1\right)\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1;x\ne1\\x\ne-1\\x\ne1\end{matrix}\right.\)

Vậy \(x\ne1;x\ne-1\) thì P được xác định.

b) \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)

\(=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)

\(=\dfrac{2x^2+x\left(x-1\right)-x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{2x^2-2x}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{2x}{x+1}\)

c) Tại x = -3 thì P được xác định nên ta có:

\(\dfrac{2x}{x+1}=\dfrac{2.\left(-3\right)}{-3+1}=\dfrac{-6}{-2}=3\)

Answer 3

b) \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\\ =\dfrac{2x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x^2+x\left(x-1\right)-x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{2x}{x+1}\)

Answer 4

a) Đkxđ: \(x\ne\pm1\)

b) Rút gọn:

\(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)

\(P=\dfrac{2x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(P=\dfrac{2x^2+x^2-x-x^2-x}{\left(x-1\right)\left(x+1\right)}\)

\(P=\dfrac{2x^2-2x}{\left(x-1\right)\left(x+1\right)}\)

\(P=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(P=\dfrac{2x}{x+1}\)

c)

- Tại \(x=-3:\)

\(P=\dfrac{2x}{x+1}=\dfrac{2.-3}{-3+1}=3\)

d)

\(\dfrac{2x}{x+1}=\dfrac{2}{x+1}+\dfrac{x}{x+1}\)

Để P có giá trị nguyên thì \(\dfrac{2}{x+1}\in Z\) .

\(\Rightarrow x+1\inƯ\left(2\right)\)

\(\Rightarrow x+1\in\left\{\pm1;\pm2\right\}\)

- \(x+1=1\Rightarrow x=0\) (tm)

- \(x+1=-1\Rightarrow x=-2\) (tm)

- \(x+1=2\Rightarrow x=1\) ( loại )

- \(x+1=-2\Rightarrow x=-3\) (tm)

Vậy có 3 giá trị của x để \(P\in Z.\)