1.x^2+1-x^4+1/x^2+1 2.5/a+1 – 1`0/a-(a^2+1) – 15/a^3+1 3.1/1-x + 1/1+x + 2/1+x^2 + 4/1+x^4 + 8/1+x^8 + 16/1+x^16 Mai kt, giúp mình với!!!!

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1. 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.$\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=\dfrac{1+x+1-x}{(1-x)(1+x)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=2.(\dfrac{1}{1-x^2}+\dfrac{1}{1+x^2})+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=4.(\dfrac{1}{1-x^4}+\dfrac{1}{1+x^4})+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}$

   $=8(\dfrac{1}{1-x^8}+\dfrac{1}{1+x^8})+\dfrac{16}{1+x^{16}}$

   $=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}$

   $=16(\dfrac{1}{1-x^{16}}+\dfrac{1}{1+x^{16}})$

   $=\dfrac{32}{1-x^{32}}$

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