Question

cho \(A=\frac{\left(x^2-2000x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}\)

tìm điều kiện để A có nghĩa

rút gọn A

Answer 1

\(\text{ĐKXĐ: }x+1\ne0\text{ và }x-2001\ne0\)

\(\Leftrightarrow x\ne-1\text{ và }x\ne2001\)

\(\frac{\left(x^2-2000x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x^2+x-2001x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}\)

\(=\frac{\left[x.\left(x+1\right)-2001\left(x+1\right)\right].2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x-2001\right)\left(x+1\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{2001}{2002}\)