Câu hỏi của Bùi Như Quỳnh

Question

$\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}$

Mashiro Shiina · Accepted Answer

$\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}$

$=\dfrac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{x^4-1}$

$=\dfrac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+1\left(x^2-1\right)}{x^4-1}$

$=\dfrac{\left(x^8+x^4+1\right)\left(x^2-1\right)}{x^4-1}$

$=\dfrac{\left(x^8+x^4+1\right)\left(x^2-1\right)}{\left(x^2+1\right)\left(x^2-1\right)}$

$=\dfrac{x^8+x^4+1}{x^2+1}$Câu trả lời: $\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}$

$=\dfrac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{x^4-1}$

$=\dfrac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+1\left(x^2-1\right)}{x^4-1}$

$=\dfrac{\left(x^8+x^4+1\right)\left(x^2-1\right)}{x^4-1}$

$=\dfrac{\left(x^8+x^4+1\right)\left(x^2-1\right)}{\left(x^2+1\right)\left(x^2-1\right)}$

$=\dfrac{x^8+x^4+1}{x^2+1}$

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