Question

Giải phương trình

\(a.\sqrt{x^2-8x+16}=x+2\)

\(b.\sqrt{x^2+6x+9}=3x-6\)

\(c.\sqrt{x^2-4x+4}-2x+5=0\)

Answer 1

Lời giải:
a. 

$\sqrt{x^2-8x+16}=x+2$

\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ x^2-8x+16=(x+2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2-8x+16=x^2+4x+4\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 12x=12\end{matrix}\right.\Leftrightarrow x=1\)

Answer 2

b.

PT \(\left\{\begin{matrix} 3x-6\geq 0\\ x^2+6x+9=(3x-6)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ x^2+6x+9=9x^2-36x+36\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 8x^2-42x+27=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ (2x-9)(4x-3)=0\end{matrix}\right.\Rightarrow x=\frac{9}{2}\)

Answer 3

c.

PT $\Leftrightarrow \sqrt{x^2-4x+4}=2x-5$
\(\Rightarrow \left\{\begin{matrix} 2x-5\geq 0\\ x^2-4x+4=(2x-5)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ x^2-4x+4=4x^2-20x+25\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ 3x^2-16x+21=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ (x-3)(3x-7)=0\end{matrix}\right.\Rightarrow x=3\)