\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-\sqrt{x+2}}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-2+2-\sqrt{x+2}}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\dfrac{4-x-2}{2+\sqrt{x+2}}}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}}{x+2}\)
\(=\dfrac{\dfrac{1}{\sqrt[3]{\left(2+6\right)^2}+2\cdot\sqrt[3]{2+6}+4}-\dfrac{1}{2+\sqrt{2+2}}}{2+2}\)
\(=\dfrac{\dfrac{1}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}-\dfrac{1}{2+2}}{4}\)
\(=\dfrac{\dfrac{1}{4+2\cdot2+4}-\dfrac{1}{4}}{4}=\left(\dfrac{1}{16}-\dfrac{1}{4}\right):4=\left(\dfrac{1}{16}-\dfrac{4}{16}\right)\cdot\dfrac{1}{4}=\dfrac{-3}{64}\)
Giúp em với ạ. Cần rất gấp ạ. Em cảm ơn ạ
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