a) (x-1)^2/x-1 + (x+1)^2/x+1 - 3
= x-1 + x+1 - 3
= 2x - 3
b) Tahy x=2 vào A ta có:
A = 2 . 2 - 3 = 1
Vậy..............
Answer:
\(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
\(ĐKXĐ:x\ne\pm1\)
\(A=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
\(=x-1+x+1-3\)
\(=2x-3\)
Thay vào ta được:
\(2.2-3\)
\(=4-3\)
\(=1\)