\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)

\(\sqrt{x^2-4}+\sqrt{x+2}=0\) (ĐK: \(x\ge2\)) 

\(\Leftrightarrow\sqrt{\left(x+2\right)\left(x-2\right)}+\sqrt{x+2}=0\) 

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}+1\right)=0\)

\(\Leftrightarrow\sqrt{x+2}=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow\text{x}=-2\left(ktm\right)\)

\(\sqrt[]{x^2-4}+\sqrt[]{x+2}=0\left(x\ge2\right)\)

\(\Leftrightarrow\sqrt[]{\left(x+2\right)\left(x-2\right)}+\sqrt[]{x+2}=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(\sqrt[]{\left(x-2\right)}+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+2}=0\\\sqrt[]{\left(x-2\right)}+1=0\left(đúng,\forall x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x\ge2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x\ge2\end{matrix}\right.\)

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

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

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

_dsaddaadad

ĐKXĐ: \(x>0\)

\(\Leftrightarrow x+\frac{1}{x}-4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+5=0\)

Đặt \(\sqrt{x}+\frac{1}{\sqrt{x}}=t>0\Rightarrow t^2=x+\frac{1}{x}+2\Rightarrow x+\frac{1}{x}=t^2-2\)

Pt trở thành:

\(t^2-2-4t+5=0\Leftrightarrow t^2-4t+3=0\) \(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{\sqrt{x}}=1\\\sqrt{x}+\frac{1}{\sqrt{x}}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}+1=0\left(vn\right)\\x-3\sqrt{x}+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)

ĐKXĐ: \(x\ge9\)

\(\Leftrightarrow\sqrt{x}+\sqrt{x-9}=\sqrt{x-1}+\sqrt{x-4}\)

\(\Leftrightarrow2x-9+2\sqrt{x^2-9x}=2x-5+2\sqrt{x^2-4x+3}\)

\(\Leftrightarrow\sqrt{x^2-9x}=2+\sqrt{x^2-4x+3}\)

Do \(x\ge9>0\Rightarrow x^2-4x>x^2-9x\Rightarrow x^2-4x+3>x^2-9x\)

\(\Rightarrow\sqrt{x^2-4x+3}+2>\sqrt{x^2-9x}\)

Pt vô nghiệm

1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

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

\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)

3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)