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Ta có:
M = \(\frac{1}{1-x}\cdot\frac{1}{1+x}\cdot\frac{1}{1+x^2}\cdot\frac{1}{1+x^4}\cdot\frac{1}{1+x^8}\cdot\frac{1}{1+x^{16}}\)
M = \(\frac{1}{\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)
M = \(\frac{1}{\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)
M = \(\frac{1}{\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)
M = \(\frac{1}{\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)
M = \(\frac{1}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
M = \(\frac{1}{1-x^{32}}\)
\(=\dfrac{1}{x+1}-\dfrac{8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-4-8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-12}{\left(x+1\right)\left(x-4\right)}=\dfrac{-11}{2\cdot\left(-3\right)}=\dfrac{11}{6}\)
\(A\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)
\(A\left(x^2+x+1\right)=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)
(Giải thích: \(\left(x^2+x+1\right)\left(x^2-x+1\right)=\left(x^2+1\right)^2-x^2=x^4+x^2+1\))
\(A\left(x^2+x+1\right)=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)...\left(x^{32}-x^{16}+1\right)\)
.....
\(A\left(x^2+x+1\right)=x^{64}-x^{32}+1\)
\(\Rightarrow A=\frac{x^{64}-x^{32}+1}{x^2+x+1}\)
\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2\left(x^2+1\right)+2.\left(x^2-1\right)}{\left(x^2+1\right)\left(1-x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2\left(x^2+1\right)+2.\left(1-x^2\right)}{\left(x^2+1\right)\left(1-x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{4\left(1+x^4\right)+4.\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(A=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(A=\frac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(x+x^{16}\right)}\)
\(A=\frac{32}{1-x^{32}}\)
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
Áp dụng (a-b) (a + b) = a 2 - b 2 . Ta có:
M = 1 1 − x 2 . 1 1 + x 2 . 1 1 + x 4 . 1 1 + x 8 . 1 1 + x 16
= 1 1 − x 16 . 1 1 + x 16 = 1 1 − x 32