a)\(x\ne\pm1\) và \(x\ne0\)
b)\(\left(\frac{x-2}{x-1}-\frac{x^2-5x-2}{x^2-1}\right).\)\(\frac{x+3}{4x}\)
\(=\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-5x-2}{\left(x-1\right)\left(x+1\right)}\right).\) \(\frac{x+3}{4x}\)
\(=\frac{x^2+x-2x-2-x^2+5x+2}{\left(x-1\right)\left(x+1\right)}.\) \(\frac{x+3}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\) \(\frac{x+3}{4x}\)
\(=\frac{4x\left(x+3\right)}{\left(x-1\right)\left(x+1\right)4x}\)
\(=\frac{x+3}{\left(x-1\right)\left(x+1\right)}\)
c)chưa nghĩ ra
\(A=\left(\frac{x-2}{x-1}-\frac{x^2-5x-2}{x^2-1}\right).\frac{x+3}{4x}\)
a) Điều kiện xác định: \(\left\{{}\begin{matrix}x-1\ne0\\x^2-1\ne0\\4x\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\\left(x-1\right).\left(x+1\right)\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne1\\x-1\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne0\end{matrix}\right.\)
Vậy để biểu thức A xác định thì \(x\ne1;x\ne-1\) và \(x\ne0.\)
b) Rút gọn:
\(A=\left(\frac{x-2}{x-1}-\frac{x^2-5x-2}{x^2-1}\right).\frac{x+3}{4x}\)
\(A=\left(\frac{x-2}{x-1}-\frac{x^2-5x-2}{\left(x-1\right).\left(x+1\right)}\right).\frac{x+3}{4x}\)
\(A=\left(\frac{\left(x-2\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{x^2-5x-2}{\left(x-1\right).\left(x+1\right)}\right).\frac{x+3}{4x}\)
\(A=\left(\frac{x^2+x-2x-2}{\left(x-1\right).\left(x+1\right)}+\frac{-\left(x^2-5x-2\right)}{\left(x-1\right).\left(x+1\right)}\right).\frac{x+3}{4x}\)
\(A=\left(\frac{x^2-x-2}{\left(x-1\right).\left(x+1\right)}+\frac{-\left(x^2-5x-2\right)}{\left(x-1\right).\left(x+1\right)}\right).\frac{x+3}{4x}\)
\(A=\left(\frac{x^2-x-2-x^2+5x+2}{\left(x-1\right).\left(x+1\right)}\right).\frac{x+3}{4x}\)
\(A=\frac{4x}{\left(x-1\right).\left(x+1\right)}.\frac{x+3}{4x}\)
\(A=\frac{4x.\left(x+3\right)}{\left(x-1\right).\left(x+1\right).4x}\)
\(A=\frac{x+3}{\left(x-1\right).\left(x+1\right)}\)
\(A=\frac{x+3}{x^2-1}.\)
Chúc bạn học tốt!