Đáp án: $-2$
Giải thích các bước giải:
Ta có:
$\lim_{x\to1}\dfrac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{(x-1)^2}$
$=\lim_{x\to1}\dfrac{(\sqrt[3]{6x-5}-(2x-1))+((2x-1)-\sqrt{4x-3})}{(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{(6x-5)-(2x-1)^3}{(\sqrt[3]{6x-5})^2+(2x-1)\sqrt[3]{6x-5}+(2x-1)^2}+\dfrac{(2x-1)^2-(4x-3)}{(2x-1)+\sqrt{4x-3}}}{(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{-4\left(x-1\right)^2\left(2x+1\right)}{(\sqrt[3]{6x-5})^2+(2x-1)\sqrt[3]{6x-5}+(2x-1)^2}+\dfrac{4(x-1)^2}{(2x-1)+\sqrt{4x-3}}}{(x-1)^2}$
$=\lim_{x\to1}\dfrac{-4\left(2x+1\right)}{(\sqrt[3]{6x-5})^2+(2x-1)\sqrt[3]{6x-5}+(2x-1)^2}+\dfrac{4}{(2x-1)+\sqrt{4x-3}}$
$=\dfrac{-4\left(2\cdot 1+1\right)}{(\sqrt[3]{6\cdot1-5})^2+(2\cdot1-1)\sqrt[3]{6\cdot1-5}+(2\cdot1-1)^2}+\dfrac{4}{(2\cdot1-1)+\sqrt{4\cdot1-3}}$
$=-2$
