a) x2 = 3x - 4
⇔ x2 - 3x + 4 = 0
Ta thấy : x2 - 3x + 4
= x2 - 2.\(\dfrac{3}{2}x+\dfrac{9}{4}+4-\dfrac{9}{4}\)
= \(\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}>0\) ∀x
⇒ PT vô nghiệm
b) x3 - 3x = 0
⇔ x( x2 - 3) = 0
⇔ x( x - \(\sqrt{3}\) )( x + \(\sqrt{3}\) ) = 0
⇔ x = 0 ; x = \(\sqrt{3}\) hoặc : x = - \(\sqrt{3}\)
KL....
\(\text{b) }x^3=3x\\ \Leftrightarrow x^3-3x=0\\ \Leftrightarrow x\left(x^2-3\right)=0\\ \Leftrightarrow x\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+\sqrt{3}=0\\x-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
Vậy pt có tập nhiệm \(S=\left\{0;-\sqrt{3};\sqrt{3}\right\}\)
\(\text{a) }x^2=3x-4\\ \Leftrightarrow x^2-3x+4=0\\ \Leftrightarrow x^2-3x+\dfrac{9}{4}+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\)
Vậy pt vô nghiệm (Vì \(\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\))
