Giải thích các bước giải:
1.$x^2+1-\dfrac{x^4-1}{x^2+1}=x^2-1-\dfrac{(x^2-1)(x^2+1)}{x^2+1}=x^2-1-(x^2-1)=0$
2.Xem lại đề
3.$\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=\dfrac{1+x+1-x}{(1-x)(1+x)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=2.(\dfrac{1}{1-x^2}+\dfrac{1}{1+x^2})+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=4.(\dfrac{1}{1-x^4}+\dfrac{1}{1+x^4})+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$
$=8(\dfrac{1}{1-x^8}+\dfrac{1}{1+x^8})+\dfrac{16}{1+x^{16}}$
$=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}$
$=16(\dfrac{1}{1-x^{16}}+\dfrac{1}{1+x^{16}})$
$=\dfrac{32}{1-x^{32}}$
