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Câu hỏi: Rút gọn phân thức:
$Q=\dfrac{x^{10}-x^{8}-x^{7}+x^{6}+x^{5}+x^{4}-x^{3}-x^{2}+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1}$
$Q=\dfrac{x^{10}-x^{8}-x^{7}+x^{6}+x^{5}+x^{4}-x^{3}-x^{2}+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1}$
Phương pháp giải
Muốn rút gọn một phân thức đại số ta làm như sau:
- Phân tích tử và mẫu thành nhân tử (nếu cần) để tìm nhân tử chung.
- Chia cả tử và mẫu cho nhân tử chung giống nhau.
Lời giải chi tiết
$\begin{aligned} Q &=\dfrac{x^{10}-x^{8}-x^{7}+x^{6}+x^{5}+x^{4}-x^{3}-x^{2}+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1} \\ &=\dfrac{\left(x^{10}-x^{8}+x^{6}\right)-\left(x^{7}-x^{5}+x^{3}\right)+\left(x^{4}-x^{2}+1\right)}{\left(x^{30}+x^{24}+x^{18}\right)+\left(x^{12}+x^{6}+1\right)} \\ &=\dfrac{x^{6}\left(x^{4}-x^{2}+1\right)-x^{3}\left(x^{4}-x^{2}+1\right)+\left(x^{4}-x^{2}+1\right)}{x^{18}\left(x^{12}+x^{6}+1\right)+\left(x^{12}+x^{6}+1\right)} \end{aligned}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{12}+x^{6}+1\right)\left(x^{18}+1\right)}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{12}+2 x^{6}+1-x^{6}\right)\left[\left(x^{6}\right)^{3}+1\right]}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left[\left(x^{12}+2 x^{6}+1\right)-x^{6}\right]\left[\left(x^{6}\right)^{3}+1\right]}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left[\left(x^{6}+1\right)^{2}-\left(x^{3}\right)^{2}\right]\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{6}+1+x^{3}\right)\left(x^{6}+1-x^{3}\right)\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{x^{4}-x^{2}+1}{\left(x^{6}+1+x^{3}\right)\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{x^{4}-x^{2}+1}{\left(x^{6}+x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{1}{\left(x^{6}+x^{3}+1\right)\left(x^{2}+1\right)\left(x^{12}-x^{6}+1\right)}$
Muốn rút gọn một phân thức đại số ta làm như sau:
- Phân tích tử và mẫu thành nhân tử (nếu cần) để tìm nhân tử chung.
- Chia cả tử và mẫu cho nhân tử chung giống nhau.
Lời giải chi tiết
$\begin{aligned} Q &=\dfrac{x^{10}-x^{8}-x^{7}+x^{6}+x^{5}+x^{4}-x^{3}-x^{2}+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1} \\ &=\dfrac{\left(x^{10}-x^{8}+x^{6}\right)-\left(x^{7}-x^{5}+x^{3}\right)+\left(x^{4}-x^{2}+1\right)}{\left(x^{30}+x^{24}+x^{18}\right)+\left(x^{12}+x^{6}+1\right)} \\ &=\dfrac{x^{6}\left(x^{4}-x^{2}+1\right)-x^{3}\left(x^{4}-x^{2}+1\right)+\left(x^{4}-x^{2}+1\right)}{x^{18}\left(x^{12}+x^{6}+1\right)+\left(x^{12}+x^{6}+1\right)} \end{aligned}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{12}+x^{6}+1\right)\left(x^{18}+1\right)}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{12}+2 x^{6}+1-x^{6}\right)\left[\left(x^{6}\right)^{3}+1\right]}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left[\left(x^{12}+2 x^{6}+1\right)-x^{6}\right]\left[\left(x^{6}\right)^{3}+1\right]}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left[\left(x^{6}+1\right)^{2}-\left(x^{3}\right)^{2}\right]\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{\left(x^{4}-x^{2}+1\right)\left(x^{6}-x^{3}+1\right)}{\left(x^{6}+1+x^{3}\right)\left(x^{6}+1-x^{3}\right)\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{x^{4}-x^{2}+1}{\left(x^{6}+1+x^{3}\right)\left(x^{6}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{x^{4}-x^{2}+1}{\left(x^{6}+x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\left(x^{12}-x^{6}+1\right)}$
$=\dfrac{1}{\left(x^{6}+x^{3}+1\right)\left(x^{2}+1\right)\left(x^{12}-x^{6}+1\right)}$