`a, (x-2)\sqrt{2x^2-x-10}=4-x^2`          (ĐKXĐ: `x <= 2, x >= 5/2`)

`⇔ (x-2)\sqrt{2x^2-x-10}+(x-2)(x+2)=0`

`⇔ (x-2)(\sqrt{2x^2-x-10}+x+2)=0`

+, Nếu `x-2=0`

`⇔ x=2`

+, Nếu `\sqrt{2x^2-x-10}+x+2=0`

`⇔ \sqrt{2x^2-x-10}=-x-2` (Đk:`x<=-2`)

`⇔ 2x^2 - x - 10=x^2-4x+4`

`⇔ x^2 + 3x - 14=0`

`⇔` $\left[\begin{matrix} x=\frac{-3+\sqrt[]{65}}{2}(loại)\\ x=x=\frac{-3-\sqrt[]{65}}{2}\end{matrix}\right.$

Vậy,..

b, `(x+3)\sqrt{x^2+3x-4}=2x^2+5x-3` (ĐKXĐ: `x<=-4, x>= 1`)

`⇔ (x+3)\sqrt{x^2+3x-4}=(x+3)(2x-1)`

`⇔ (x+3)(\sqrt{x^2+3x-4}-2x+1)=0`

+, Nếu `x+3=0`

`⇔ x=-3`

+, Nếu `\sqrt{x^2+3x-4}-2x+1=0`

`⇔ \sqrt{x^2+3x-4}=2x-1` (ĐK: `x>=1/2`)

`⇔ x^2+3x-4=4x^2-4x+1`

`⇔ 3x^2 -7x + 5=0` (vô nghiệm)

Vậy,..

`color{purple}{@SHIN}`