Question

Giải phương trình

a) \(\sqrt{x^2-2x+4}=2x-2\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\)

c) \(\sqrt{2x^2-2x+1}=2x-1\)

d) \(\sqrt{x+4\sqrt{x-4}}=2\)

Answer 1

a. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)

b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$

$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2$

$\Leftrightarrow |\sqrt{x-1}+1|=2$

$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$

$\Leftrightarrow x=2$ (tm)

Answer 2

c. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)

d.

ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$

$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$

$\Leftrightarrow \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

Answer 3

a: Ta có: \(\sqrt{x^2-2x+4}=2x-2\)

\(\Leftrightarrow x^2-2x+4=4x^2-8x+4\)

\(\Leftrightarrow-3x^2+6x=0\)

\(\Leftrightarrow-3x\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

b: Ta có: \(\sqrt{x+2\sqrt{x-1}}=2\)

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)

\(\Leftrightarrow\sqrt{x-1}+1=2\)

\(\Leftrightarrow x-1=1\)

hay x=2

c: Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)

\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)

\(\Leftrightarrow-2x^2+2x=0\)

\(\Leftrightarrow-2x\left(x-1\right)=0\)

hay x=1

Answer 4

d:Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

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

\(\Leftrightarrow x-4=0\)

hay x=4