Question

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

a, tim dieu kien xac dinh

b, rut gon A

Answer 1

 `a, x^2 +x` \(\ne\) `0` \(\Leftrightarrow x\left(x+1\right)\ne0\)

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

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

`b, A=(x+1)/(x^2+x) =(x+1)/(x(x+1))=1/x`

Answer 2

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

\(a,\) Điều kiện xác định: \(x^2+x\ne0\Leftrightarrow x\left(x+1\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

\(b,A=\dfrac{x+1}{x^2+x}=\dfrac{x+1}{x\left(x+1\right)}=\dfrac{1}{x}\)