Toán P=3√x+2 phần 2√x-1 cộng với √x-1 phần √x+4 trừ cho x-6√x+5 phần 2x+7vx-4 01/07/2021 By Alexandra P=3√x+2 phần 2√x-1 cộng với √x-1 phần √x+4 trừ cho x-6√x+5 phần 2x+7vx-4
Đáp án: \(\begin{array}{l}a)\dfrac{{4\sqrt x + 1}}{{2\sqrt x – 1}}\\b)Min = 3\\c)x = 0\end{array}\) Giải thích các bước giải: \(\begin{array}{l}a)DK:x \ge 0;x \ne \left\{ {\dfrac{1}{4};1} \right\}\\P = \dfrac{{3\sqrt x + 2}}{{2\sqrt x – 1}} + \dfrac{{\sqrt x – 1}}{{\sqrt x + 4}} – \dfrac{{x – 6\sqrt x + 5}}{{2x + 7\sqrt x – 4}}\\ = \dfrac{{\left( {3\sqrt x + 2} \right)\left( {\sqrt x + 4} \right) + \left( {\sqrt x – 1} \right)\left( {2\sqrt x – 1} \right) – x + 6\sqrt x – 5}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\ = \dfrac{{3x + 14\sqrt x + 8 + 2x – 3\sqrt x + 1 – x + 6\sqrt x – 5}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\ = \dfrac{{4x + 17\sqrt x + 4}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\ = \dfrac{{\left( {4\sqrt x + 1} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\ = \dfrac{{4\sqrt x + 1}}{{2\sqrt x – 1}}\\b)P = \dfrac{{2\left( {\sqrt x + 1} \right) – 1}}{{2\sqrt x – 1}} = 2 – \dfrac{1}{{2\sqrt x – 1}}\\Do:2\sqrt x \ge 0\forall x \ge 0\\ \to 2\sqrt x – 1 \ge – 1\\ \to \dfrac{1}{{2\sqrt x – 1}} \le – 1\\ \to – \dfrac{1}{{2\sqrt x – 1}} \ge 1\\ \to 2 – \dfrac{1}{{2\sqrt x – 1}} \ge 3\\ \to Min = 3\\ \Leftrightarrow x = 0\\c)P \in Z \to \dfrac{1}{{2\sqrt x – 1}} \in Z\\ \to 2\sqrt x – 1 \in U\left( 1 \right)\\ \to \left[ \begin{array}{l}2\sqrt x – 1 = 1\\2\sqrt x – 1 = – 1\end{array} \right.\\ \to \left[ \begin{array}{l}x = 1\left( l \right)\\x = 0\end{array} \right.\end{array}\) Trả lời
Đáp án:
\(\begin{array}{l}
a)\dfrac{{4\sqrt x + 1}}{{2\sqrt x – 1}}\\
b)Min = 3\\
c)x = 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne \left\{ {\dfrac{1}{4};1} \right\}\\
P = \dfrac{{3\sqrt x + 2}}{{2\sqrt x – 1}} + \dfrac{{\sqrt x – 1}}{{\sqrt x + 4}} – \dfrac{{x – 6\sqrt x + 5}}{{2x + 7\sqrt x – 4}}\\
= \dfrac{{\left( {3\sqrt x + 2} \right)\left( {\sqrt x + 4} \right) + \left( {\sqrt x – 1} \right)\left( {2\sqrt x – 1} \right) – x + 6\sqrt x – 5}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\
= \dfrac{{3x + 14\sqrt x + 8 + 2x – 3\sqrt x + 1 – x + 6\sqrt x – 5}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\
= \dfrac{{4x + 17\sqrt x + 4}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\
= \dfrac{{\left( {4\sqrt x + 1} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x + 4} \right)\left( {2\sqrt x – 1} \right)}}\\
= \dfrac{{4\sqrt x + 1}}{{2\sqrt x – 1}}\\
b)P = \dfrac{{2\left( {\sqrt x + 1} \right) – 1}}{{2\sqrt x – 1}} = 2 – \dfrac{1}{{2\sqrt x – 1}}\\
Do:2\sqrt x \ge 0\forall x \ge 0\\
\to 2\sqrt x – 1 \ge – 1\\
\to \dfrac{1}{{2\sqrt x – 1}} \le – 1\\
\to – \dfrac{1}{{2\sqrt x – 1}} \ge 1\\
\to 2 – \dfrac{1}{{2\sqrt x – 1}} \ge 3\\
\to Min = 3\\
\Leftrightarrow x = 0\\
c)P \in Z \to \dfrac{1}{{2\sqrt x – 1}} \in Z\\
\to 2\sqrt x – 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
2\sqrt x – 1 = 1\\
2\sqrt x – 1 = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)