Question

Giải các phương trình :

a) \(\sqrt{3x-4}=x-3\)

b) \(\sqrt{x^2-2x+3}=2x-1\)

c) \(\sqrt{2x^2+3x+7}=x+2\)

d) \(\sqrt{3x^2-4x-4}=\sqrt{2x+5}\)

Answer 1

a)
Pt\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=\left(x-3\right)^2\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4=x^2-6x+9\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+13=0\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{9+\sqrt{29}}{2}\\x_2=\dfrac{9-\sqrt{29}}{2}\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{9+\sqrt{29}}{2}\)
Vậy \(x=\dfrac{9+\sqrt{29}}{2}\) là nghiệm của phương trình.

Answer 2

b) Pt \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+3=\left(2x-1\right)^2\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-2=0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{7}}{3}\\x_2=\dfrac{1-\sqrt{7}}{3}\end{matrix}\right.\\x\ge\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{7}}{3}\)
Vậy phương trình có duy nhất nghiệm là: \(x=\dfrac{1+\sqrt{7}}{3}\)

Answer 3

c) Pt \(\Leftrightarrow\left\{{}\begin{matrix}2x^2+3x+7=\left(x+2\right)^2\\x+2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+3=0\\x\ge-2\end{matrix}\right.\) (vô nghiệm)

Answer 4

d) pt\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-4x-4=2x+5\\2x+5\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-6x-9=0\\x\ge\dfrac{-5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=-1\\x_3=3\end{matrix}\right.\\x\ge\dfrac{-5}{2}\end{matrix}\right.\)\(\Leftrightarrow x=3\).
Vậy x = 3 là nghiệm của phương trình.