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\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^5+1}-\frac{16}{x^{16}+1}\)
\(=\frac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2\left(x^2+1\right)-2.\left(x^2-1\right)}{x^2-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2x^2+2-2x^2+2}{\left(x^2+1\right)\left(x^2-1\right)}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8.\left(x^8+1\right)-8\left(x^8-1\right)}{\left(x^8-1\right)\left(x^8+1\right)}-\frac{16}{x^{16}+1}\)
\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}\)
\(=\frac{16.\left(x^{16}+1\right)-16.\left(x^{16}-1\right)}{\left(x^{16}-1\right)\left(x^{16}+1\right)}\)
\(=\frac{32}{x^{32}-1}\)
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,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
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 *-*
\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2\left(x^2+1\right)+2.\left(x^2-1\right)}{\left(x^2+1\right)\left(1-x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{2\left(x^2+1\right)+2.\left(1-x^2\right)}{\left(x^2+1\right)\left(1-x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{4\left(1+x^4\right)+4.\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(A=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(A=\frac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(x+x^{16}\right)}\)
\(A=\frac{32}{1-x^{32}}\)
Theo đầu bài ta có:
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(1+x\right)+\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(2+2x^2\right)+\left(2-2x^2\right)}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(4+4x^4\right)+\left(4-4x^4\right)}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(8+8x^8\right)+\left(8-8x^8\right)}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)+\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{\left(16+16x^{16}\right)+\left(1-x^{16}\right)}{1-x^{32}}\)
\(=\frac{17+15x^{16}}{1-x^{32}}\)
a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)
Quy đồng :
\(\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2x}{x^2-x+1}=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x+1}=\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)
\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
c ) MTC : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{32}{1-x^{32}}\)