giúp e với mn em đang cần gấp á 1h30 em phải đi rồi giúp em với November 17, 2020 by Cherry giúp e với mn em đang cần gấp á 1h30 em phải đi rồi giúp em với
Đáp án: a. \(\dfrac{{4x}}{{\sqrt x – 3}}\) Giải thích các bước giải: \(\begin{array}{l}a.DK:x > 0;x \ne 4\\P = \left[ {\dfrac{{4\sqrt x \left( {2 – \sqrt x } \right) + 8x}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right]:\left[ {\dfrac{{\sqrt x – 1 – 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 2} \right)}}} \right]\\ = \dfrac{{8\sqrt x – 4x + 8x}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{ – \sqrt x \left( {2 – \sqrt x } \right)}}{{ – \sqrt x + 3}}\\ = \dfrac{{4x + 8\sqrt x }}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 – \sqrt x } \right)}}{{\sqrt x – 3}}\\ = \dfrac{{4\sqrt x .\sqrt x }}{{\sqrt x – 3}}\\ = \dfrac{{4x}}{{\sqrt x – 3}}\\b.P = – 1\\ \to \dfrac{{4x}}{{\sqrt x – 3}} = – 1\\ \to 4x = – \sqrt x + 3\\ \to 4x + \sqrt x – 3 = 0\\ \to \left[ \begin{array}{l}\sqrt x = \dfrac{3}{4}\\\sqrt x = – 1\left( l \right)\end{array} \right.\\ \to x = \dfrac{9}{{16}}\\c.m\left( {\sqrt x – 3} \right).\dfrac{{4x}}{{\sqrt x – 3}} > x + 1\\ \to 4xm > x + 1\\ \to x + 1 – 4xm < 0\\ \to x\left( {1 – 4m} \right) + 1 < 0\\ \to x < – \dfrac{1}{{1 – 4m}}\\Do:x > 9\\ \to – \dfrac{1}{{1 – 4m}} > 9\\ \to \dfrac{1}{{1 – 4m}} < – 9\\ \to \dfrac{{1 + 9 – 36m}}{{1 – 4m}} < 0\\ \to \dfrac{{10 – 36m}}{{1 – 4m}} < 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}10 – 36m > 0\\1 – 4m < 0\end{array} \right.\\\left\{ \begin{array}{l}10 – 36m < 0\\1 – 4m > 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}\dfrac{5}{{18}} > m\\m > \dfrac{1}{4}\end{array} \right.\\\left\{ \begin{array}{l}\dfrac{5}{{18}} < m\\m < \dfrac{1}{4}\end{array} \right.\left( l \right)\end{array} \right.\end{array}\) Reply
Đáp án:
a. \(\dfrac{{4x}}{{\sqrt x – 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 4\\
P = \left[ {\dfrac{{4\sqrt x \left( {2 – \sqrt x } \right) + 8x}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right]:\left[ {\dfrac{{\sqrt x – 1 – 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 2} \right)}}} \right]\\
= \dfrac{{8\sqrt x – 4x + 8x}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{ – \sqrt x \left( {2 – \sqrt x } \right)}}{{ – \sqrt x + 3}}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 – \sqrt x } \right)}}{{\sqrt x – 3}}\\
= \dfrac{{4\sqrt x .\sqrt x }}{{\sqrt x – 3}}\\
= \dfrac{{4x}}{{\sqrt x – 3}}\\
b.P = – 1\\
\to \dfrac{{4x}}{{\sqrt x – 3}} = – 1\\
\to 4x = – \sqrt x + 3\\
\to 4x + \sqrt x – 3 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{3}{4}\\
\sqrt x = – 1\left( l \right)
\end{array} \right.\\
\to x = \dfrac{9}{{16}}\\
c.m\left( {\sqrt x – 3} \right).\dfrac{{4x}}{{\sqrt x – 3}} > x + 1\\
\to 4xm > x + 1\\
\to x + 1 – 4xm < 0\\
\to x\left( {1 – 4m} \right) + 1 < 0\\
\to x < – \dfrac{1}{{1 – 4m}}\\
Do:x > 9\\
\to – \dfrac{1}{{1 – 4m}} > 9\\
\to \dfrac{1}{{1 – 4m}} < – 9\\
\to \dfrac{{1 + 9 – 36m}}{{1 – 4m}} < 0\\
\to \dfrac{{10 – 36m}}{{1 – 4m}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
10 – 36m > 0\\
1 – 4m < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
10 – 36m < 0\\
1 – 4m > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{5}{{18}} > m\\
m > \dfrac{1}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{5}{{18}} < m\\
m < \dfrac{1}{4}
\end{array} \right.\left( l \right)
\end{array} \right.
\end{array}\)