Ta có: \(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{x+1-\left(x-1\right)}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

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

\(=\frac{2\left(x^2+1\right)-2\left(x^2-1\right)}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2x^2+2-2x^2+2}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4x^4+4-4x^4+4}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

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

\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}=\frac{16x^{16}+16-16x^{16}+16}{\left(x^{16}-1\right)\left(x^{16}+1\right)}=\frac{32}{x^{32}-1}\)

a,     (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))

=  (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\))  \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\)  - \(\dfrac{21}{60}\))

= - \(\dfrac{3}{80}\)  \(\times\) (- \(\dfrac{2}{3}\))

= \(\dfrac{1}{40}\) 

b, (-1)³ + (- \(\dfrac{2}{3}\))² : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)

=  -1³ +   \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1

= 0

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

=

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

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

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

=\(\dfrac{32}{1-x^{32}}\)

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

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

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

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

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

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

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)

a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)

b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)

c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)

khó quá làm sao mà trả lời đc

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