\(\text { Tính giới hạn } A=\lim _{x \rightarrow 2} \frac{\sqrt{5-2 x}-2 \sqrt{x-1}+2 x-3}{\sqrt{2 x-3}+\sqrt{6 x-3}-2 x} \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } A=\lim \limits _{x \rightarrow 2}=\frac{(\sqrt{5-2 x}-1)-2(\sqrt{x-1}-1)+2(x-2)}{(\sqrt{2 x-3}-1)+(\sqrt{6 x-3}-3)-2(x-2)} \\ &=\lim \limits _{x \rightarrow 2} \frac{-\frac{2(x-2)}{\sqrt{5-2 x}+1}-\frac{2(x-2)}{\sqrt{x-1}+1}+2(x-2)}{\frac{2(x-2)}{\sqrt{2 x-3}+1}+\frac{6(x-2)}{\sqrt{6 x-3}+3}-2(x-2)}=\lim \limits _{x \rightarrow 2} \frac{(x-2)\left(-\frac{2}{\sqrt{5-2 x}+1}-\frac{2}{\sqrt{x-1}+1}+2\right)}{(x-2)\left(\frac{2}{\sqrt{2 x-3}+1}+\frac{6}{\sqrt{6 x-3}+3}-2\right)} \\ &=\lim \limits _{x \rightarrow 2} \frac{1-\frac{2}{\sqrt{5-2 x}+1}+1-\frac{2}{\sqrt{x-1}+1}}{\frac{2}{\sqrt{2 x-3}+1}-1+\frac{6}{\sqrt{6 x-3}+3}-1}=\lim \limits _{x \rightarrow 2} \frac{\frac{\sqrt{5-2 x}-1}{\sqrt{5-2 x}+1}+\frac{\sqrt{x-1}-1}{\sqrt{x-1}+1}}{\frac{1-\sqrt{2 x-3}}{\sqrt{2 x-3}+1}+\frac{3-\sqrt{6 x-3}}{\sqrt{6 x-3}+3}} \\ &=\lim \limits _{x \rightarrow 2} \frac{-\frac{2}{(\sqrt{5-2 x}+1)^{2}}+\frac{x-2}{(\sqrt{x-1}+1)^{2}}}{-\frac{2(x-2)}{(\sqrt{2 x-3}+1)^{2}}-\frac{6(x-2)}{(\sqrt{6 x-3}+3)^{2}}}=\lim \limits _{x \rightarrow 2} \frac{-\frac{2}{(\sqrt{5-2 x}+1)^{2}}+\frac{1}{(\sqrt{x-1}+1)^{2}}}{-\frac{2}{(\sqrt{2 x-3}+1)^{2}}-\frac{6}{(\sqrt{6 x-3}+3)^{2}}}=\frac{3}{8} . \end{aligned} \)