\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

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

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

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

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

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

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

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

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

=>\(4\sqrt{x-1}=x+3\)

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

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

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

điều kiện xác đinh \(x\ge-\frac{1}{2}\)

ta có \(x\left(5x^3+2\right)-2\left(\sqrt{2x+1}-1\right)=0\)

\(\Leftrightarrow5x^4+2x-2\sqrt{2x+1}+2=0\Leftrightarrow5x^4+2x+1-2\sqrt{2x+1}+1=0\)

\(\Leftrightarrow5x^4+\left(\sqrt{2x+1}-1\right)^2=0=>\orbr{\begin{cases}5x^4=0\\\sqrt{2x+1}-1=0\end{cases}\Leftrightarrow x=0\left(nhận\right)}\)

zậy \(S=\left\{0\right\}\)

ĐK: \(x\ge\frac{-1}{2}\). PT đã cho có thể viết lại thành 

\(5x^4+\left(\sqrt{2x+1}-1\right)^2=0\)

Do \(5x^4\ge0,\left(\sqrt{2x+1}-1\right)^2\ge0\)nên PT trên chỉ thỏa mãn khi \(\hept{\begin{cases}5x^4=0\\\left(\sqrt{2x+1}-1\right)^2=0\end{cases}}\)

Giải hệ này ta được x=0

Vậy PT đã cho có nghiệm duy nhất x=0

ĐKXĐ: \(\hept{\begin{cases}x^2-5x+2\ge0\\2x-1>0\\x-2\ge0\end{cases}\Leftrightarrow x\ge2}\)

Phương trình 

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

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

<=> 2x-1=x² hoặc \(\hept{\begin{cases}3-x\ge0\\x-2=3-x^2\end{cases}}\)

<=> x²-2x+1=0 hoặc \(\hept{\begin{cases}x\le3\\x^2-7x+11=0\end{cases}}\)

<=> x=1 hoặc \(\hept{\begin{cases}x\le3\\x=\frac{7\pm\sqrt{3}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{7-\sqrt{5}}{2}\end{cases}}\)

Đối chiếu điều kiện x>=2 => x=\(=\frac{7-\sqrt{5}}{2}\left(tm\right)\)

Vậy pt có nghiệm \(x=\frac{7-\sqrt{5}}{2}\)

đề bài có sai ko vậy