## Nhang nha cần gấp. Toán 9 rút gọn.

Question

Nhang nha cần gấp.
Toán 9 rút gọn.

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4 years 2020-10-21T07:23:35+00:00 2 Answers 70 views 0

1. Đáp án:

Giải thích các bước giải:

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2. Đáp án:

b. $$\dfrac{1}{{\sqrt x }}$$

Giải thích các bước giải:

$$\begin{array}{l} a.DK:x \ge 0;x \ne 9\\ \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) – 3x + 1 – \sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{2x – 6\sqrt x + x + 4\sqrt x + 3 – 3x + 1 – \sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{ – 3\sqrt x + 4}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\ b.DK:x > 0;x \ne 1\\ \left[ {\dfrac{{x – \sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x – 1 + 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\ = \dfrac{1}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\ = \dfrac{1}{{\sqrt x }}\\ c.DK:a > 0;a \ne 1\\ \left[ {\dfrac{{\sqrt a – \sqrt a + 1}}{{\sqrt a \left( {\sqrt a – 1} \right)}}} \right]:\left( {\dfrac{{\sqrt a + 1 – \sqrt a – 2}}{{\sqrt a – 1}}} \right)\\ = \dfrac{1}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\sqrt a – 1}}{{ – 1}}\\ = – \dfrac{1}{{\sqrt a }} \end{array}$$