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10) Выполняем деление "столбиком", как показано в первом вложении. Ответ: x²-x+1
![\displaystyle 11) \ \frac{x^6-1}{3x^3-3x}= \frac{(x^3-1)(x^3+1)}{3x(x^2-1)} = \\ \frac{(x-1)(x^2+x+1)(x+1)(x^2-x+1)}{3x(x-1)(x+1)}= \qquad |x| \neq 1 \\ \frac{(x^2+x+1)(x^2-x+1)}{3x}=\frac{[(x^2+1)+x][(x^2+1)-x]}{3x}= \\ \frac{(x^2+1)^2-x^2}{3x}= \frac{x^4+2x^2+1-x^2}{3x}= \frac{x^4+x^2+1}{3x}](https://tex.z-dn.net/?f=%5Cdisplaystyle++11%29+%5C+%5Cfrac%7Bx%5E6-1%7D%7B3x%5E3-3x%7D%3D++%5Cfrac%7B%28x%5E3-1%29%28x%5E3%2B1%29%7D%7B3x%28x%5E2-1%29%7D+%3D+%5C%5C++%5Cfrac%7B%28x-1%29%28x%5E2%2Bx%2B1%29%28x%2B1%29%28x%5E2-x%2B1%29%7D%7B3x%28x-1%29%28x%2B1%29%7D%3D+%5Cqquad+%7Cx%7C+%5Cneq+1+%5C%5C+%5Cfrac%7B%28x%5E2%2Bx%2B1%29%28x%5E2-x%2B1%29%7D%7B3x%7D%3D%5Cfrac%7B%5B%28x%5E2%2B1%29%2Bx%5D%5B%28x%5E2%2B1%29-x%5D%7D%7B3x%7D%3D+%5C%5C++%5Cfrac%7B%28x%5E2%2B1%29%5E2-x%5E2%7D%7B3x%7D%3D+%5Cfrac%7Bx%5E4%2B2x%5E2%2B1-x%5E2%7D%7B3x%7D%3D+%5Cfrac%7Bx%5E4%2Bx%5E2%2B1%7D%7B3x%7D+ "\displaystyle 11) \ \frac{x^6-1}{3x^3-3x}= \frac{(x^3-1)(x^3+1)}{3x(x^2-1)} = \\ \frac{(x-1)(x^2+x+1)(x+1)(x^2-x+1)}{3x(x-1)(x+1)}= \qquad |x| \neq 1 \\ \frac{(x^2+x+1)(x^2-x+1)}{3x}=\frac{[(x^2+1)+x][(x^2+1)-x]}{3x}= \\ \frac{(x^2+1)^2-x^2}{3x}= \frac{x^4+2x^2+1-x^2}{3x}= \frac{x^4+x^2+1}{3x}")
