Đáp án: $\dfrac45$
Giải thích các bước giải:
Đặt $\sqrt[4]{x}=t\to x=t^4$
$\to I=\lim_{x\to1}\dfrac{2x^2-3x+1}{x\sqrt[4]{x}-1}$ trở thành
$I =\lim_{t\to1}\dfrac{2t^8-3t^4+1}{t^5-1}$
$\to I=\lim_{x\to1}\dfrac{\left(t+1\right)\left(t-1\right)\left(t^2+1\right)\left(\sqrt[4]{2}t+1\right)\left(\sqrt[4]{2}t-1\right)\left(\sqrt{2}t^2+1\right)}{\left(t-1\right)\left(t^4+t^3+t^2+t+1\right)}$
$\to I=\dfrac{\left(t+1\right)\left(t^2+1\right)\left(\sqrt[4]{2}t+1\right)\left(\sqrt[4]{2}t-1\right)\left(\sqrt{2}t^2+1\right)}{t^4+t^3+t^2+t+1}$
$\to I=\dfrac{\left(1+1\right)\left(1^2+1\right)\left(\sqrt[4]{2}\cdot \:1+1\right)\left(\sqrt[4]{2}\cdot \:1-1\right)\left(\sqrt{2}\cdot \:1^2+1\right)}{1^4+1^3+1^2+1+1}$
$\to I=\dfrac45$
