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Môn Toán Calantha 4 years 2020-11-11T03:55:01+00:00 2020-11-11T03:55:01+00:00 2 Answers 76 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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Delwyn · Answer 1 · 2020-11-11T03:56:10+00:00

Đáp án:

1) $A = \frac{3}{5}$

Giải thích các bước giải:

$\begin{array}{l} 1) T h a y : x = 16 \\ \to A = \frac{\sqrt{16} - 1}{\sqrt{16} + 1} = \frac{3}{5} \\ 2) B = \frac{x + 2}{(\sqrt{x} + 1) (x - \sqrt{x} + 1)} + \frac{\sqrt{x}}{x - \sqrt{x} + 1} - \frac{1}{\sqrt{x} + 1} \\ = \frac{x + 2 + \sqrt{x} (\sqrt{x} + 1) - x + \sqrt{x} - 1}{(\sqrt{x} + 1) (x - \sqrt{x} + 1)} \\ = \frac{x + 2 + x + \sqrt{x} - x + \sqrt{x} - 1}{(\sqrt{x} + 1) (x - \sqrt{x} + 1)} \\ = \frac{x + 2 \sqrt{x} + 1}{(\sqrt{x} + 1) (x - \sqrt{x} + 1)} \\ = \frac{{(\sqrt{x} + 1)}^{2}}{(\sqrt{x} + 1) (x - \sqrt{x} + 1)} \\ = \frac{\sqrt{x} + 1}{x - \sqrt{x} + 1} \\ \to d p c m \\ 3) A . B < \frac{1}{2} \\ \to \frac{\sqrt{x} - 1}{\sqrt{x} + 1} . \frac{\sqrt{x} + 1}{x - \sqrt{x} + 1} < \frac{1}{2} \\ \to \frac{\sqrt{x} - 1}{x - \sqrt{x} + 1} < \frac{1}{2} \\ \to \frac{2 \sqrt{x} - 2 - x + \sqrt{x} - 1}{2 (x - \sqrt{x} + 1)} < 0 \\ \to \frac{- x + 3 \sqrt{x} - 3}{2 (x - \sqrt{x} + 1)} < 0 \\ D o : {\begin{matrix} - x + 3 \sqrt{x} - 3 < 0 \forall x \geq 0 \\ x - \sqrt{x} + 1 > 0 \forall x \geq 0 \end{matrix} \\ \to \frac{- x + 3 \sqrt{x} - 3}{2 (x - \sqrt{x} + 1)} < 0 (l d) \\ K L : x \geq 0 \end{array}$

giahung · Answer 2 · 2020-11-11T03:56:38+00:00

giahung

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2020-11-11T03:56:38+00:00 November 11, 2020 at 3:56 am

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Đáp án:

1) $A = \frac{3}{5}$

Giải thích các bước giải:

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