Question

Tìm x:

\(\sqrt{x^2-2x+4}=2x-3\)

Answer 1

Lời giải:
PT \(\Rightarrow \left\{\begin{matrix} 2x-3\geq 0\\ x^2-2x+4=(2x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{3}{2}\\ 3x^2-10x+5=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{3}{2}\\ x=\frac{5\pm \sqrt{10}}{3}\end{matrix}\right.\Leftrightarrow x=\frac{5+\sqrt{10}}{3}\)

Answer 2

\(\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-60=40>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{6}\\x=\dfrac{5+\sqrt{10}}{6}\end{matrix}\right.\)

Answer 3

\(\sqrt{x^2-2x+4}=2x-3\)

\(\Leftrightarrow x^2-2x+4=4x^2-12x+9\)

\(\Leftrightarrow4x^2-12x+9-x^2+2x-4=0\)

\(\Leftrightarrow3x^2-10x+5=0\)

\(\text{Δ}=\left(-10\right)^2-4\cdot3\cdot5=100-60=40\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{10-2\sqrt{5}}{6}=\dfrac{5-\sqrt{5}}{3}\\x_2=\dfrac{10+2\sqrt{5}}{6}=\dfrac{5+\sqrt{5}}{3}\end{matrix}\right.\)