\lim_(x->1)(\sqrt(2x-1)-\root(3)(3x^(2)-3x+1))/((x-1)^(2)) 27/10/2021 Bởi aihong \lim_(x->1)(\sqrt(2x-1)-\root(3)(3x^(2)-3x+1))/((x-1)^(2))
Đáp án: Giải thích các bước giải: \[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{2x-1}-\sqrt[3]{3{{x}^{2}}-3x+1}}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{(\sqrt{2x-1}-x)+(x-\sqrt[3]{3{{x}^{2}}-3x+1})}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{2x-1}-x}{{{(x-1)}^{2}}}+\frac{x-\sqrt[3]{3{{x}^{2}}-3x+1}}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{-{{x}^{2}}+2x-1}{{{(x-1)}^{2}}(\sqrt{2x-1}+x)}+\frac{{{x}^{3}}-3{{x}^{2}}+3x-1}{{{(x-1)}^{2}}({{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}})} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{{{-(x-1)}^{2}}}{{{(x-1)}^{2}}(\sqrt{2x-1}+x)}+\frac{{{(x-1)}^{3}}}{{{(x-1)}^{2}}({{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}})} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{-1}{\sqrt{2x-1}+x}+\frac{x-1}{{{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}}} \\ & =\frac{-1}{\sqrt{2-1}+1}+\frac{1-1}{{{1}^{2}}+\sqrt[3]{3-3+1}+\sqrt[3]{{{(3-3+1)}^{2}}}}=-\frac{1}{2} \\ \end{align}\] Bình luận
Đáp án: Giải thích các bước giải: $ \dfrac{\sqrt{2x – 1} – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$ $ = \dfrac{\sqrt{2x – 1} – x + x – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$ $ = \dfrac{\sqrt{2x – 1} – x}{(x – 1)²} + \dfrac{x – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$ $ = \dfrac{2x – 1 – x²}{(x – 1)²(\sqrt{2x – 1} + x)} + \dfrac{x³ – (3x² – 3x + 1)}{(x – 1)²(x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²})}$ $ = \dfrac{- (1 – x)²}{(x – 1)²(\sqrt{2x – 1} + x)} + \dfrac{(x – 1)³}{(x – 1)²(x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²})}$ $ = \dfrac{- 1}{\sqrt{2x – 1} + x} + \dfrac{x – 1}{x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²}}$ Vậy: $ \lim_{x \to 1} \dfrac{\sqrt{2x – 1} – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$ $ = \lim_{x \to 1}\dfrac{- 1}{\sqrt{2x – 1} + x} + \lim_{x \to 1}\dfrac{x – 1}{x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²}}$ $ = \dfrac{-1}{\sqrt{2.1 – 1} + 1} + \dfrac{1 – 1}{1² + 1.\sqrt[3]{3.1² – 3.1 + 1} + \sqrt[3]{(3.1² – 3.1 + 1)²}} $ $ = – \dfrac{1}{2} + 0 = – \dfrac{1}{2}$ Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{2x-1}-\sqrt[3]{3{{x}^{2}}-3x+1}}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{(\sqrt{2x-1}-x)+(x-\sqrt[3]{3{{x}^{2}}-3x+1})}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{2x-1}-x}{{{(x-1)}^{2}}}+\frac{x-\sqrt[3]{3{{x}^{2}}-3x+1}}{{{(x-1)}^{2}}} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{-{{x}^{2}}+2x-1}{{{(x-1)}^{2}}(\sqrt{2x-1}+x)}+\frac{{{x}^{3}}-3{{x}^{2}}+3x-1}{{{(x-1)}^{2}}({{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}})} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{{{-(x-1)}^{2}}}{{{(x-1)}^{2}}(\sqrt{2x-1}+x)}+\frac{{{(x-1)}^{3}}}{{{(x-1)}^{2}}({{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}})} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{-1}{\sqrt{2x-1}+x}+\frac{x-1}{{{x}^{2}}+x\sqrt[3]{3{{x}^{2}}-3x+1}+\sqrt[3]{{{(3{{x}^{2}}-3x+1)}^{2}}}} \\ & =\frac{-1}{\sqrt{2-1}+1}+\frac{1-1}{{{1}^{2}}+\sqrt[3]{3-3+1}+\sqrt[3]{{{(3-3+1)}^{2}}}}=-\frac{1}{2} \\ \end{align}\]
Đáp án:
Giải thích các bước giải:
$ \dfrac{\sqrt{2x – 1} – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$
$ = \dfrac{\sqrt{2x – 1} – x + x – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$
$ = \dfrac{\sqrt{2x – 1} – x}{(x – 1)²} + \dfrac{x – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$
$ = \dfrac{2x – 1 – x²}{(x – 1)²(\sqrt{2x – 1} + x)} + \dfrac{x³ – (3x² – 3x + 1)}{(x – 1)²(x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²})}$
$ = \dfrac{- (1 – x)²}{(x – 1)²(\sqrt{2x – 1} + x)} + \dfrac{(x – 1)³}{(x – 1)²(x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²})}$
$ = \dfrac{- 1}{\sqrt{2x – 1} + x} + \dfrac{x – 1}{x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²}}$
Vậy:
$ \lim_{x \to 1} \dfrac{\sqrt{2x – 1} – \sqrt[3]{3x² – 3x + 1}}{(x – 1)²}$
$ = \lim_{x \to 1}\dfrac{- 1}{\sqrt{2x – 1} + x} + \lim_{x \to 1}\dfrac{x – 1}{x² + x\sqrt[3]{3x² – 3x + 1} + \sqrt[3]{(3x² – 3x + 1)²}}$
$ = \dfrac{-1}{\sqrt{2.1 – 1} + 1} + \dfrac{1 – 1}{1² + 1.\sqrt[3]{3.1² – 3.1 + 1} + \sqrt[3]{(3.1² – 3.1 + 1)²}} $
$ = – \dfrac{1}{2} + 0 = – \dfrac{1}{2}$