giúp em với mọi ngườiiiiiiiiiii mai em cần huhu em cảm ơn ạ November 3, 2020 by Maris giúp em với mọi ngườiiiiiiiiiii mai em cần huhu em cảm ơn ạ
Đáp án: a. \(\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\) Giải thích các bước giải: \(\begin{array}{l}a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\M = \dfrac{{2\sqrt x – 9 + \left( {2\sqrt x + 1} \right)\left( {\sqrt x – 2} \right) – \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{{2\sqrt x – 9 + 2x – 3\sqrt x – 2 – x + 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{{x – \sqrt x – 2}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\b.M = 5\\ \to \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = 5\\ \to \sqrt x + 1 = 5\sqrt x – 15\\ \to 4\sqrt x = 16\\ \to \sqrt x = 4\\ \to x = 16\left( {TM} \right)\\c.M = \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = \dfrac{{\sqrt x – 3 + 4}}{{\sqrt x – 3}} = 1 + \dfrac{4}{{\sqrt x – 3}}\\M \in Z\\ \to \dfrac{4}{{\sqrt x – 3}} \in Z\\ \to \sqrt x – 3 \in U\left( 4 \right)\\ \to \left[ \begin{array}{l}\sqrt x – 3 = 4\\\sqrt x – 3 = – 4\left( l \right)\\\sqrt x – 3 = 2\\\sqrt x – 3 = – 2\\\sqrt x – 3 = 1\\\sqrt x – 3 = – 1\end{array} \right. \to \left[ \begin{array}{l}\sqrt x = 7\\\sqrt x = 5\\\sqrt x = 1\\\sqrt x = 4\\\sqrt x = 2\end{array} \right.\\ \to \left[ \begin{array}{l}x = 49\\x = 25\\x = 1\\x = 16\\x = 4(l)\end{array} \right.\end{array}\) ( Đề đoạn cuối \(\sqrt {x – 3} \) t sửa thành \({\sqrt x – 3}\) mới làm được nha ) Reply
Đáp án:
a. \(\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
M = \dfrac{{2\sqrt x – 9 + \left( {2\sqrt x + 1} \right)\left( {\sqrt x – 2} \right) – \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{2\sqrt x – 9 + 2x – 3\sqrt x – 2 – x + 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{x – \sqrt x – 2}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
b.M = 5\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = 5\\
\to \sqrt x + 1 = 5\sqrt x – 15\\
\to 4\sqrt x = 16\\
\to \sqrt x = 4\\
\to x = 16\left( {TM} \right)\\
c.M = \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = \dfrac{{\sqrt x – 3 + 4}}{{\sqrt x – 3}} = 1 + \dfrac{4}{{\sqrt x – 3}}\\
M \in Z\\
\to \dfrac{4}{{\sqrt x – 3}} \in Z\\
\to \sqrt x – 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x – 3 = 4\\
\sqrt x – 3 = – 4\left( l \right)\\
\sqrt x – 3 = 2\\
\sqrt x – 3 = – 2\\
\sqrt x – 3 = 1\\
\sqrt x – 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 7\\
\sqrt x = 5\\
\sqrt x = 1\\
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 49\\
x = 25\\
x = 1\\
x = 16\\
x = 4(l)
\end{array} \right.
\end{array}\)
( Đề đoạn cuối \(\sqrt {x – 3} \) t sửa thành \({\sqrt x – 3}\) mới làm được nha )