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Ta có:\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{2}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\frac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{16+16}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
Theo đầu bài ta có:
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(1+x\right)+\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(2+2x^2\right)+\left(2-2x^2\right)}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(4+4x^4\right)+\left(4-4x^4\right)}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(8+8x^8\right)+\left(8-8x^8\right)}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)+\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{\left(16+16x^{16}\right)+\left(1-x^{16}\right)}{1-x^{32}}\)
\(=\frac{17+15x^{16}}{1-x^{32}}\)
bài 1
\(ĐKXĐ:1+x\ne0\Rightarrow x\ne-1\)
\(\frac{3-7x}{1+x}=\frac{1}{2}\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\\
\Leftrightarrow-14x-x=1-6\\
\Leftrightarrow-15x=-5\\
\Leftrightarrow x=\frac{1}{3}\left(N\right)\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
\(a,\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2-6x}\)
\(=\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x\left(x-6\right)}\)
\(=\frac{7\left(x-6\right)\left(x+6\right)-x\left(x-6\right)+36\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)}\)
\(=\frac{7\left(x^2-6\right)-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{7x^2-42-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{6x^2+42x+216}{x\left(x^2-6\right)}\)
\(=\frac{6\left(x^2+7x+36\right)}{x\left(x^2-6\right)}\)
Đề sai nhé, phải là như này nè :
\(b,\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-x^2+x-1}{x^4+x^2+1}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x}{x^4+x^2+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x\left(x^4-x^2+1\right)-2x\left(x^4+x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x^5-2x^3+2x-2x^5-2x^3-2x}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=-\frac{4x^3}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}=0\)
\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)
\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}\)
\(A=\frac{16}{1-x^{16}}\)