7881 câu Trắc nghiệm tổng hợp môn Toán 2023 cực hay có đáp án (Phần 76)

Rút gọn biểu thức: A = (x - 2 căn bậc hai x) / (x căn bậc hai x - 1) + (căn bậc hai x + 1)

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Rút gọn biểu thức:

\[A = \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{{x^2} - \sqrt x }}\] (x > 0, x ¹ 1)

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Giải thích

\[A = \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{{x^2} - \sqrt x }}\] (x > 0, x ¹ 1)

\[ = \frac{{x - 2\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x\sqrt x + x + \sqrt x }} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {x\sqrt x - 1} \right)}}\]

\[ = \frac{{x - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\sqrt x + 1}}{{\sqrt x \left( {x + \sqrt x + 1} \right)}} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{\sqrt x \left( {x - 2\sqrt x } \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{x\sqrt x - 2x + x - 1 + 1 + 2x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{x\sqrt x + x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{\sqrt x \left( {x + \sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\]

\[ = \frac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\]

Vậy \[A = \frac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\].