0 thoughts on “Giúp em bài 3 câu 3 với ạ. Em đg cần gấp”
Đáp án:
\(Min = 2\sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l} Q = \dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right) + 5\sqrt x – 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{x – 3\sqrt x + 2 + 5\sqrt x – 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \dfrac{{\sqrt x }}{{\sqrt x – 2}}\\ \dfrac{P}{Q} = \dfrac{{x + 3}}{{\sqrt x – 2}}:\dfrac{{\sqrt x }}{{\sqrt x – 2}}\\ = \dfrac{{x + 3}}{{\sqrt x }} = \sqrt x + \dfrac{3}{{\sqrt x }}\\ Do:x > 0\\ \to BDT:Co – si:\sqrt x + \dfrac{3}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{3}{{\sqrt x }}} \\ \to \sqrt x + \dfrac{3}{{\sqrt x }} \ge 2\sqrt 3 \\ \to Min = 2\sqrt 3 \\ \Leftrightarrow \sqrt x = \dfrac{3}{{\sqrt x }}\\ \to x = 3 \end{array}\)
Đáp án:
\(Min = 2\sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
Q = \dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right) + 5\sqrt x – 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x – 3\sqrt x + 2 + 5\sqrt x – 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x – 2}}\\
\dfrac{P}{Q} = \dfrac{{x + 3}}{{\sqrt x – 2}}:\dfrac{{\sqrt x }}{{\sqrt x – 2}}\\
= \dfrac{{x + 3}}{{\sqrt x }} = \sqrt x + \dfrac{3}{{\sqrt x }}\\
Do:x > 0\\
\to BDT:Co – si:\sqrt x + \dfrac{3}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{3}{{\sqrt x }}} \\
\to \sqrt x + \dfrac{3}{{\sqrt x }} \ge 2\sqrt 3 \\
\to Min = 2\sqrt 3 \\
\Leftrightarrow \sqrt x = \dfrac{3}{{\sqrt x }}\\
\to x = 3
\end{array}\)