Cho P= ($\frac{2+\sqrt{x}}{2-\sqrt{x}}$ – $\frac{2-\sqrt{x}}{2+\sqrt{x}}$ – $\frac{4x}{x-4}$ ) : $\frac{\sqrt{x}-3}{2\sqrt{x}-x}$ a)Rút gọn b)Tìm x để

Question

Cho P= ($\frac{2+\sqrt{x}}{2-\sqrt{x}}$ – $\frac{2-\sqrt{x}}{2+\sqrt{x}}$ – $\frac{4x}{x-4}$ ) : $\frac{\sqrt{x}-3}{2\sqrt{x}-x}$
a)Rút gọn
b)Tìm x để \P\ =1

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3 tuần 2021-08-21T17:12:19+00:00 2 Answers 0 views 0

1. CHÚC BẠN HỌC TỐT!!!

Giải thích các bước giải:

2. Giải thích các bước giải:

a,

ĐKXĐ: $$\left\{ \begin{array}{l} x > 0\\ x \ne 4\\ x \ne 9 \end{array} \right.$$

Ta có:

$$\begin{array}{l} P = \left( {\dfrac{{2 + \sqrt x }}{{2 – \sqrt x }} – \dfrac{{2 – \sqrt x }}{{2 + \sqrt x }} – \dfrac{{4x}}{{x – 4}}} \right):\dfrac{{\sqrt x – 3}}{{2\sqrt x – x}}\\ = \left( {\dfrac{{2 + \sqrt x }}{{2 – \sqrt x }} – \dfrac{{2 – \sqrt x }}{{2 + \sqrt x }} – \dfrac{{4x}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\dfrac{{\sqrt x – 3}}{{\sqrt x .\left( {2 – \sqrt x } \right)}}\\ = \dfrac{{{{\left( {2 + \sqrt x } \right)}^2} – {{\left( {2 – \sqrt x } \right)}^2} + 4x}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x – 3}}{{\sqrt x .\left( {2 – \sqrt x } \right)}}\\ = \dfrac{{\left( {x + 4\sqrt x + 4} \right) – \left( {x – 4\sqrt x + 4} \right) + 4x}}{{\left( {2 – \sqrt x } \right).\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x .\left( {2 – \sqrt x } \right)}}{{\sqrt x – 3}}\\ = \dfrac{{4x + 8\sqrt x }}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x .\left( {2 – \sqrt x } \right)}}{{\sqrt x – 3}}\\ = \dfrac{{4\sqrt x \left( {\sqrt x + 2} \right)}}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x – 3}}\\ = \dfrac{{4\sqrt x .\sqrt x }}{{\sqrt x – 3}}\\ = \dfrac{{4x}}{{\sqrt x – 3}}\\ b,\\ \left| P \right| = 1 \Leftrightarrow \left[ \begin{array}{l} P = 1\\ P = – 1 \end{array} \right.\\ TH1:\,\,\,P = 1 \Leftrightarrow \dfrac{{4x}}{{\sqrt x – 3}} = 1\\ \Leftrightarrow 4x = \sqrt x – 3\\ \Leftrightarrow 4x – \sqrt x + 3 = 0\\ \Leftrightarrow {\left( {2\sqrt x } \right)^2} – 2.2\sqrt x .\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{47}}{{16}} = 0\\ \Leftrightarrow {\left( {2\sqrt x – \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} = 0\,\,\,\,\,\left( {vn} \right)\\ TH2:\,\,\,P = – 1 \Leftrightarrow \dfrac{{4x}}{{\sqrt x – 3}} = – 1\\ \Leftrightarrow 4x = – \sqrt x + 3\\ \Leftrightarrow 4x + \sqrt x – 3 = 0\\ \Leftrightarrow \left( {\sqrt x + 1} \right)\left( {4\sqrt x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt x = – 1\\ \sqrt x = \dfrac{3}{4} \end{array} \right. \Leftrightarrow \sqrt x = \dfrac{3}{4} \Leftrightarrow x = \dfrac{9}{{16}} \end{array}$$