Lời giải:

`x^2+2sqrt[2x+7]=2sqrt[-2x+3]+5 (x in NN^[***]; -7/2<=x<=3/2)`

`<=> x^2+(2sqrt[2x+7]-6)=(2sqrt[-2x+3]-2)+1`

`<=> x^2+2(sqrt[2x+7]-3)-2(sqrt[-2x+3]-1)-1=0`

`<=> x^2+2 * [2x+7-9]/[sqrt[2x+7]+3]-2 * [-2x+3-1]/[sqrt[-2x+3]+1]-1=0`

`<=> x^2+[2(2x-2)]/[sqrt[2x+7]+3]-[2(-2x+2)]/[sqrt[-2x+3]+1]-1=0`

`<=> (x-1)(x+1)+[4(x-1)]/[sqrt[2x+7]+3]-[-4(x-1)]/[sqrt[-2x+3]+1]=0`

`<=> (x-1)(x+1+4/[sqrt[2x+7]+3]+4/[sqrt[-2x+3]+1])=0`

TH1 : `x-1=0`

`<=> x=1` (nhận)

TH2 : `x+1+4/[sqrt[2x+7]+3]+4/[sqrt[-2x+3]+1]=0`

Với điều kiện `-7/2<=x<=3/2` thì `x+1+4/[sqrt[2x+7]+3]+4/[sqrt[-2x+3]+1]>0` 

`=>` Vô nghiệm

Vậy `x=1` , tức là Cá chỉ yêu mỗi mình T

\(x^2+2\sqrt{2x+7}=2\sqrt{-2x+3}+5\\Dk:x\in\left[-\dfrac{7}{2},\dfrac{3}{2}\right]\\\Rightarrow x=1\quad \qquad \\pt\Leftrightarrow x+5-2\sqrt{2x+7}+2\sqrt{-2x+3}+x-3-x^2-2x+3=0\\\Leftrightarrow \dfrac{(x+5)^2-4(2x+7)}{x+5+2\sqrt{2x+7}}+\dfrac{4(-2x+3)-(x-3)^2}{x-3-2\sqrt{-2x+3}}-x^2-2x+3=0\\\Leftrightarrow \dfrac{x^2+2x-3}{x+5+2\sqrt{2x+7}}+\dfrac{x^2+2x-3}{x-3-2\sqrt{-2x+3}}-x^2-2x+3=0\\\Leftrightarrow (x^2+2x-3)\left[ \dfrac{1}{x+5+2\sqrt{2x+7}}+\dfrac{1}{x-3-2\sqrt{-2x+3}}+1\right]=0\\\dfrac{1}{x+5+2\sqrt{2x+7}}+\dfrac{1}{x-3-2\sqrt{-2x+3}}\ge \dfrac{2}{x+1+\sqrt{2x+7}-\sqrt{-2x+3}}> 0\\\Rightarrow\left[ \begin{array}{l}x=1\\x=-3(\text{loai})\end{array} \right.\)