a) \(x\ne0\) , \(x\ne-1\) , \(x\ne1\)
b)
\(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{1-2x+x^2}\)
\(=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{\left(x-1\right)^2}\)
\(=\left(\dfrac{1-\left(2-x\right).x}{x\left(x+1\right)}\right).\dfrac{3x}{\left(x-1\right)^2}\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}.\dfrac{3x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^2.3x}{x\left(x+1\right)\left(x-1\right)^2}\)
\(=\dfrac{3x}{x\left(x+1\right)}=\dfrac{3}{x+1}\)
Với x =5 , ta có :
\(A=\dfrac{3}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)
Với x =0, ta có ;
\(A=\dfrac{3}{0+1}=3\)
Vậy x = 5 \(\Leftrightarrow\) \(A=\dfrac{1}{2}\)
\(x=0\Leftrightarrow A=3\)
a)ĐKXĐ:
\(\left\{{}\begin{matrix}x^2+x\ne0\\x+1\ne0\\1-2x+x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0;x\ne-1\\x\ne-1\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne-1\end{matrix}\right.\)
Vậy...
b)Với \(x\ne0;x\ne1;x\ne-1\)thì:
\(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{1-2x+x^2}\\ A=\left[\dfrac{1}{x\left(x+1\right)}-\dfrac{2-x}{x+1}\right].\dfrac{3x}{\left(x-1\right)^2}\\ A=\dfrac{1 -2x+x^2}{x\left(x+1\right)}.\dfrac{3x}{\left(x-1\right)^2}\\ A=\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}.\dfrac{3x}{\left(x-1\right)^2}\\ A=\dfrac{3}{x+1}\)
Vậy...
c) Với \(x\ne0;x\ne1,x\ne-1\)
+)x=5(tm) Khi đó PT có dạng:
\(A=\dfrac{3}{5+1}\\ A=\dfrac{1}{2}\)
+)x=0(không tm đkxđ)
Vậy...
\(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{1-2x+x^2}\)
\(=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{2}{x+1}\right).\dfrac{3x}{\left(x-1\right)^2}\)
( điều kiện ở đây là xét mẫu của phân số đều phải khác 0 hết)
( nhớ là điều kiện nào trùng chỉ lấy 1)
\(\left\{{}\begin{matrix}x\left(x+1\right)\ne0&x+1\ne0&\left(x-1\right)^2\ne0&&\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\\x\ne1\end{matrix}\right.\)
