x6 - 7x3 - 8 = 0
\(\Leftrightarrow\left(x^3+1\right)\left(x^3-8\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x-2\right)\left(x^2+2x+4\right)=0\left(1\right)\)Do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) và \(x^2+2x+4=\left(x+1\right)^2+3>0\) với mọi x
Nên (1) \(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x\in\left\{-1;2\right\}\)
\(\left(x^3-1\right)\left(x^3+8\right)=0\)
\(\left\{{}\begin{matrix}x^3-1=0\Rightarrow x=1\\x^3+8=0\Rightarrow x=-2\end{matrix}\right.\)
cách này mình thấy dễ hơn , bạn tham khảo cũng được nhhé !!
đặt t=x^3
ta có:
t^2 - 7t -8= 0
t^2 - 8t + t - 8 =0
(t^2 - 8t) +( t - 8 ) = 0
t( t-8) + ( t - 8) = 0
( t+ 1) ( t-8) = 0
\(\left[{}\begin{matrix}t-8\\t+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=8\\t=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^3=8\\x^3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
