Question

1.Giải pt : 2x⁴ + 5x² - 7 = 0

Answer 1

\(2x^4+5x^2-7=0\left(1\right)\)

Đặt \(t=x^2\left(t\ge0\right)\)

\(\left(1\right):2t^2+5t-7=0\\ \Leftrightarrow2t^2+7t-2t-7=0\\ \Leftrightarrow t\left(2t+7\right)-\left(2t+7\right)=0\\ \Leftrightarrow\left(2t+7\right)\left(t-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2t+7=0\\t-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{7}{2}\left(KTM\right)\\t=1\left(TM\right)\end{matrix}\right.\)

Với \(t=1\Leftrightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy nghiệm phương trình là \(S=\left\{1;-1\right\}\)