a: \(\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x+5}-\sqrt{x^2+x-4}}{-x^2+x+12}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x^2+x-4}-\sqrt{x+5}}{x^2-x-12}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{x^2+x-4-x-5}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{\left(x^2-9\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}\cdot\dfrac{1}{\left(x-4\right)\left(x+3\right)}\right)\)\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{x-3}{\left(x-4\right)\left(\sqrt{x^2+x-4}+\sqrt{x+5}\right)}\right)\)\(=\dfrac{-3-3}{\left(-3-4\right)\left(\sqrt{9-3-4}+\sqrt{-3+5}\right)}\)\(=\dfrac{-6}{\left(-7\right)\left(\sqrt{2}+\sqrt{2}\right)}=\dfrac{6}{7\cdot2\sqrt{2}}=\dfrac{3}{7\sqrt{2}}=\dfrac{3\sqrt{2}}{14}\)b: \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\cdot\left(\sqrt{x+1}-1\right)+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\dfrac{x}{\sqrt{x+1}+1}\right)+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\left(\sqrt[3]{8-x}\right)^2}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{x+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)\(=\dfrac{2}{\sqrt{0+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)\(=\dfrac{2}{1+1}+\dfrac{1}{4+2\cdot2+4}\)\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)c: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)\(=\lim\limits_{x\rightarrow2}\dfrac{x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-1\right)\left(x-3\right)}=\dfrac{2}{\left(2-1\right)\left(2-3\right)}=-2\) Câu trả lời: a: \(\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x+5}-\sqrt{x^2+x-4}}{-x^2+x+12}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x^2+x-4}-\sqrt{x+5}}{x^2-x-12}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{x^2+x-4-x-5}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{\left(x^2-9\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}\cdot\dfrac{1}{\left(x-4\right)\left(x+3\right)}\right)\)\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{x-3}{\left(x-4\right)\left(\sqrt{x^2+x-4}+\sqrt{x+5}\right)}\right)\)\(=\dfrac{-3-3}{\left(-3-4\right)\left(\sqrt{9-3-4}+\sqrt{-3+5}\right)}\)\(=\dfrac{-6}{\left(-7\right)\left(\sqrt{2}+\sqrt{2}\right)}=\dfrac{6}{7\cdot2\sqrt{2}}=\dfrac{3}{7\sqrt{2}}=\dfrac{3\sqrt{2}}{14}\)b: \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\cdot\left(\sqrt{x+1}-1\right)+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\dfrac{x}{\sqrt{x+1}+1}\right)+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\left(\sqrt[3]{8-x}\right)^2}}{x}\)\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{x+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)\(=\dfrac{2}{\sqrt{0+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)\(=\dfrac{2}{1+1}+\dfrac{1}{4+2\cdot2+4}\)\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)c: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)\(=\lim\limits_{x\rightarrow2}\dfrac{x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-1\right)\left(x-3\right)}=\dfrac{2}{\left(2-1\right)\left(2-3\right)}=-2\)

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22 tháng 11 2023

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#Toán lớp 11

Nguyễn Lê Phước Thịnh

22 tháng 11 2023

a: \(\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x+5}-\sqrt{x^2+x-4}}{-x^2+x+12}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x^2+x-4}-\sqrt{x+5}}{x^2-x-12}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{x^2+x-4-x-5}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{\left(x^2-9\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)

\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}\cdot\dfrac{1}{\left(x-4\right)\left(x+3\right)}\right)\)

\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{x-3}{\left(x-4\right)\left(\sqrt{x^2+x-4}+\sqrt{x+5}\right)}\right)\)

\(=\dfrac{-3-3}{\left(-3-4\right)\left(\sqrt{9-3-4}+\sqrt{-3+5}\right)}\)

\(=\dfrac{-6}{\left(-7\right)\left(\sqrt{2}+\sqrt{2}\right)}=\dfrac{6}{7\cdot2\sqrt{2}}=\dfrac{3}{7\sqrt{2}}=\dfrac{3\sqrt{2}}{14}\)

b: \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\cdot\left(\sqrt{x+1}-1\right)+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\dfrac{x}{\sqrt{x+1}+1}\right)+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\left(\sqrt[3]{8-x}\right)^2}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{x+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)

\(=\dfrac{2}{\sqrt{0+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)

\(=\dfrac{2}{1+1}+\dfrac{1}{4+2\cdot2+4}\)

\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)

c: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-1\right)\left(x-3\right)}=\dfrac{2}{\left(2-1\right)\left(2-3\right)}=-2\)

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