$ĐKXĐ : x ≥ 1$
Ta có pt : $4x\sqrt[]{x-1} = 4x^2+x-10$
$⇔ 4x\sqrt[]{x-1} - 8 = 4x^2+x-18$
$⇔\dfrac{16x^2.(x-1)-64}{4x\sqrt[]{x-1} + 8} = (x-2).(4x+9)$
$⇔ \dfrac{16x^3-16x^2-64}{4x\sqrt[]{x-1} + 8} - (x-2).(4x+9) = 0 $
$⇔ \dfrac{16.(x^3-x^2-4)}{4.(x\sqrt[]{x-1} + 2)} -(x-2).(4x+9) =0$
$⇔ \dfrac{4.(x^3-x^2-4)}{x\sqrt[]{x-1} + 2} -(x-2).(4x+9) = 0 $
$⇔ \dfrac{4.(x-2).(x^2+x+2)}{x\sqrt[]{x-1} + 2} - (x-2).(4x=9) = 0 $
$⇔ (x-2).\bigg[\dfrac{4.(x^2+x+2)}{x\sqrt[]{x-1}+2} - 4x-9\bigg] = 0$ $(*)$
Do $x\sqrt[]{x-1} + 2 ≥ 2 > 0 ∀ x \in ĐKXĐ$
$x^2+x+2 = \bigg(x+\dfrac{1}{2}\bigg)^2 + \dfrac{7}{4} ≥ \dfrac{7}{4}>0$
$\to 4.(x^2+x+2) ≥ 7 > 0 $
Do đó : $\dfrac{4.(x^2+x+2)}{x\sqrt[]{x-1}+2} ≤\dfrac{7}{2}$
$\to \dfrac{4.(x^2+x+2)}{x\sqrt[]{x-1}+2} - 9 < 0 $
Với $x≥1$ thì $-4x < 0 $
Nên : $\dfrac{4.(x^2+x+2)}{x\sqrt[]{x-1}}+2 - 4x-9 < 0 ∀ x \in ĐKXĐ$
Từ $(*)$ suy ra $x-2=0 ⇔x=2$ ( Thỏa mãn )
Vậy pt có nghiệm duy nhất $x=2$
