1. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$

$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$

$\Leftrightarrow 22=10\sqrt{x-4}$

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

$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$

(thỏa mãn)

2. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$

$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$

$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$

$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)

3. ĐKXĐ: $x\geq 3$

Bình phương 2 vế thu được:

$3x-2+2\sqrt{(2x+1)(x-3)}=4x$
$\Leftrightarrow 2\sqrt{(2x+1)(x-3)}=x+2$

$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$

$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$

$\Leftrightarrow (x-4)(7x+4)=0$

Do $x\geq 3$ nên $x=4$

Thử lại thấy thỏa mãn

Vậy $x=4$

4. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$

$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$

Ta thấy, với mọi $x\geq 4$ thì:

$(\sqrt{x}-2)^2\ge 0$

$2021\sqrt{x-4}\geq 0$ 

Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

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

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

1.

ĐKXĐ: \(x\ge3\)

Đặt \(\sqrt{x-3}=t\ge0\Rightarrow x=t^2+3\)

Pt trở thành:

\(t^2+3-7t-9=0\)

\(\Leftrightarrow t^2-7t-6=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{7-\sqrt{73}}{2}< 0\left(loại\right)\\t=\dfrac{7+\sqrt{73}}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-3}=\dfrac{7+\sqrt{73}}{2}\)

\(\Rightarrow x=\dfrac{67+7\sqrt{73}}{2}\)

Nghiệm xấu quá, em nói giáo viên ra đề kiểm tra lại đề là \(x-7\sqrt{x-3}-9=0\) hay \(x-7\sqrt{x-3}+9=0\) nhé

2.

ĐKXĐ: \(x\ge2\)

\(\sqrt{x+3}+\sqrt{x-2}=5\)

\(\Leftrightarrow2x+1+2\sqrt{\left(x+3\right)\left(x-2\right)}=25\)

\(\Leftrightarrow\sqrt{x^2+x-6}=12-x\) (\(x\le12\))

\(\Rightarrow x^2+x-6=\left(12-x\right)^2\)

\(\Leftrightarrow x^2+x-6=144-24x+x^2\)

\(\Rightarrow x=6\)

Cách 2:

\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{x-2}-2=0\)

\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)

\(\Leftrightarrow x=6\)

3.

ĐKXĐ: \(x\ge8+8\sqrt{2}\)

Đặt \(\sqrt{x+4}=t>0\) \(\Rightarrow x=t^2-4\)

Pt trở thành:

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

\(\Leftrightarrow t^2-4t-4=9\)

\(\Leftrightarrow t^2-4t-13=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{17}\\t=2-\sqrt{17}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+4}=2+\sqrt{17}\)

\(\Leftrightarrow x=17+4\sqrt{17}\)

Như câu 1, em nhờ giáo viên ra đề kiểm tra lại là \(\sqrt{x-4\sqrt{x+4}}=3\) hay \(\sqrt{x-4\sqrt{x-4}}=3\)

2. ĐKXĐ: $x\geq 2$

PT \(\Rightarrow x+3=(5-\sqrt{x-2})^2\)

\(\Leftrightarrow x+3=25+x-2-10\sqrt{x-2}\)

\(\Leftrightarrow 20=10\sqrt{x-2}\Leftrightarrow x-2=4\Leftrightarrow x=6\)

Thử lại thấy thỏa mãn

Vậy $x=6$

3. ĐKXĐ: $x\geq -4$

PT $\Leftrightarrow \sqrt{(x+4)-4\sqrt{x+4}+4}=3$

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

$\Leftrightarrow |\sqrt{x+4}-2|=3$

$\Leftrightarrow \sqrt{x+4}-2=\pm 3$. TH $\sqrt{x+4}-2=-3$ loại vì $\sqrt{x+4}-2\geq -2> -3$

Do đó: $\sqrt{x+4}-2=3$

$\Leftrightarrow \sqrt{x+4}=5$

$\Leftrightarrow x+4=25$

$\Leftrightarrow x=21$ (thỏa mãn)

Vậy $x=21$

** Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.

Lời giải:

1. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow x-9=7\sqrt{x-3}$

\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ (x-9)^2=49(x-3)\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ x^2-67x+228=0\end{matrix}\right.\Rightarrow x=\frac{67+7\sqrt{73}}{2}\)

a: ĐKXĐ: x>=1

\(\dfrac{1}{2}\sqrt{x-1}-\sqrt{4x-4}+3=0\)

=>\(3+\dfrac{1}{2}\sqrt{x-1}-2\sqrt{x-1}=0\)

=>\(3-\dfrac{3}{2}\sqrt{x-1}=0\)

=>\(\dfrac{3}{2}\sqrt{x-1}=3\)

=>\(\sqrt{x-1}=2\)

=>x-1=4

=>x=5(nhận)

b: \(\sqrt{x^2-4x+4}+x-2=0\)

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

=>|x-2|=-(x-2)

=>x-2<=0

=>x<=2

c: 

ĐKXĐ: 7-x>=0

=>x<=7

\(\sqrt{7-x}+1=x\)

=>\(\sqrt{7-x}=x-1\)

=>\(\left\{{}\begin{matrix}x-1>=0\\7-x=x^2-2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1< =x< =7\\x^2-2x+1-7+x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1< =x< =7\\x^2-x-6=0\end{matrix}\right.\Leftrightarrow x=3\)