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

=>căn x-3=0

=>x-3=0

=>x=3

2: =>\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot4+16}=5\)

=>\(\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
=>2*căn 2x-3+5=5

=>2x-3=0

=>x=3/2

Điều kiện: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}=0}\)

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

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

Với \(\dfrac{1}{2}\le x< 1\)

\(\Leftrightarrow1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=0\)

\(\Leftrightarrow-2\sqrt{2x-1}+6=0\)

\(\Leftrightarrow x=5\left(l\right)\)

Tương tự cho các trường hợp: \(1\le x< \dfrac{5}{2};\dfrac{5}{2}\le x< 5;x\ge5\)

Tới đây thì kết luận thôi.

\(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}}=0\)

ĐK:\(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}-2\sqrt{2x-1-4\sqrt{2x-1}+4}+3\sqrt{2x-1-6\sqrt{2x-1}+9}=0\)

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

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

\(\Leftrightarrow\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=0\)

\(\Leftrightarrow2\sqrt{2x-1}-6=0\)\(\Leftrightarrow\sqrt{2x-1}=3\)

\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Rightarrow x=5\) *Thỏa*

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)

\(\sqrt[3]{2x+1}+\sqrt[3]{2x+2}+\sqrt[3]{2x+3}=0\)

\(\Rightarrow\left(2x+1\right)+\left(2x+2\right)+\left(2x+3\right)=3\sqrt[3]{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)}\)

\(\Leftrightarrow6x+6-3\sqrt[3]{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)}=0\)

\(\Leftrightarrow2x+2-\sqrt[3]{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)}=0\)

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

Trường hợp 1:

\(\sqrt[3]{2x+2}=0\)

\(\Leftrightarrow x=-1\)

Trường hợp 2:

\(\sqrt[3]{\left(2x+2\right)^2}-\sqrt[3]{\left(2x+1\right)\left(2x+3\right)}=0\)

\(\Leftrightarrow\left(2x+2\right)^2-\left(2x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow0x+1=0\)

Pt vô n_o

Vậy phương trình có 1 nghiệm duy nhất . . .

\(\sqrt[3]{2x+1}+\sqrt[3]{2x+2}+\sqrt[3]{2x+3}=0\left(1\right)\)

Pt (1) <=> \(\sqrt[3]{2x+1}+\sqrt[3]{2x+2}=-\sqrt[3]{2x+3}\) (*)

\(\Leftrightarrow\left(\sqrt[3]{2x+1}+\sqrt[3]{2x+2}\right)^3=-\left(2x+3\right)\)

\(\Leftrightarrow4x+3+3\sqrt[3]{2x+1}\cdot\sqrt[3]{2x+2}\cdot\left(\sqrt[3]{2x+1}+\sqrt[3]{2x+2}\right)=-\left(2x+3\right)\) (2)

Thay (*) vào (2) ta được:

\(\left(2\right)\Leftrightarrow\sqrt[3]{2x+1}\cdot\sqrt[3]{2x+2}\cdot\sqrt[3]{2x+3}=-2x-2\)

\(\Leftrightarrow\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)=\left(-2x-2\right)^3\)

\(\Leftrightarrow\left(2x+2\right)\cdot\left[\left(2x+2\right)\left(2x+3\right)+\left(2x+2\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\8x^2+18x+10=0\end{matrix}\right.\)

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

Thử lại chỉ có x = -1 thỏa mãn

Vậy pt có 1 nghiệm duy nhất là \(x=-1\)