\(PT\Leftrightarrow x^6+4x^3-5=0\)
\(\Leftrightarrow\left(x^3-1\right)\left(x^3+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\sqrt[3]{5}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\sqrt[3]{5}\right\}\)
\(x^3(x^3 + 4) = 5\) \(↔x^6 + 4x^3 – 5 = 0\)
\( ↔x^6 - x^3 + 5x^3 – 5 = 0\)
\(↔(x^3 – 1)x^3 + 5(x^3 – 1) = 0\)
\( ↔(x^3 – 1)(x^3 + 5) = 0\)
\(\leftrightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+5=0\end{matrix}\right.\)
\(\leftrightarrow\left[{}\begin{matrix}x=\sqrt[3]{1}=1\\x=\sqrt[3]{-5}\end{matrix}\right.\)
➤ \(x\in\left\{1;\sqrt[3]{-5}\right\}\)
\(\Leftrightarrow x^6+4x^3-5=0\\ \Leftrightarrow x^6+5x^3-x^3-5=0\\ \Leftrightarrow\left(x^6+5x^3\right)-\left(x^3+5\right)=0\\ \Leftrightarrow x^3\left(x^3+5\right)-\left(x^3+5\right)=0\\ \Leftrightarrow\left(x^3-1\right)\left(x^3+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^3=1\\x^3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\sqrt[3]{5}\end{matrix}\right.\)
\(x^3\left(x^3+4\right)=5\)
=> \(\left(x^3\right)^2+4x^3-5=0\)
Đặt x3 lak a , ta có :
\(a^2+4a-5=0\)
=> \(a^2+4a+4-9=0\)
=> \(\left(a+2\right)^2-3^2=0\)
=> \(\left(a+2-3\right)\left(a+2+3\right)=0\)
=> \(\left[{}\begin{matrix}a-1=0\\a+5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=1\\a=-5\end{matrix}\right.\)
Hay : \(\left[{}\begin{matrix}x^3=1\\x^3=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\-\sqrt[3]{5}\end{matrix}\right.\)
Vậy tập nghiệm của pt S = ...........
