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\(=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right)+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$
a, \(\frac{x^{32}+x^{16}+1}{x^{16}+x^8+1}\)
\(=\frac{x^8+x^4+1}{x^4+x^2+1}\) Vậy phân thức \(a=\frac{x^8+x^4+1}{x^4+x^2+1}\)
P/s; Căn thức a, là phân số tối giản
b, \(\frac{x^8+3x^4+4}{x^4+x^2+2}\)
\(=\frac{x^4+3x^2+2}{x^2+x^1+1}\) Vậy căn thức \(b=\frac{x^4+3x^2+2}{x^2+x^1+1}\)
P/s; Căn thức b, có thể rút gọn được cho 2 và 4
Em ko chắc đâu nhé *-*
Ta có:\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{2}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\frac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{16+16}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{32}{1-x^{32}}\)
