\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x-1-\sqrt{4x-3}\right)+\left(\sqrt[3]{6x-5}-\left(2x-1\right)\right)}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{4\left(x-1\right)^2}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(x-1\right)^2\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{4}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}\right)=-2\)
\(lim\left(x->1\right)\dfrac{-\sqrt{4x-3}+\sqrt[3]{6x-5}}{\left(x-1\right)^2}\)
Đặt \(\sqrt{4x-3}=f\left(x\right);\sqrt[3]{6x-5}=g\left(x\right)\Rightarrow g\left(x\right)^6-f\left(x\right)^6=4\left(x-1\right)^2\left(16x-13\right)\)
\(f\left(1\right)=1;g\left(1\right)=1\)
Ta có
\(lim\left(x->1\right)\dfrac{-f\left(x\right)+g\left(x\right)}{\left(x-1\right)^2}=lim\left(x->1\right)\dfrac{g\left(x\right)^6-f\left(x\right)^6}{\left(x-1\right)^2}\cdot\dfrac{1}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)
\(=lim\left(x->1\right)\dfrac{4\left(16x-3\right)}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)
\(=\dfrac{4\left(16-3\right)}{1^5+1^4\cdot1+1^3\cdot1^2+1^2\cdot1^3+1\cdot1^4+1^5}=\dfrac{26}{3}\)
