bn đăng từng bài 1 thôi ạ
Bài 3
a) ĐK \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Thay \(x=9\) vào Q ta đc \(Q=\dfrac{\sqrt{9}+2}{\sqrt{9}-2}=\dfrac{3+2}{3-2}=5\)
b) \(P=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x-2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c) \(M=\dfrac{P}{Q}=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(M< \dfrac{1}{2}< =>\dfrac{\sqrt{x}}{\sqrt{x}+2}< \dfrac{1}{2}< =>2\sqrt{x}< \sqrt{x}+2\)
\(< =>\sqrt{x}< 2< =>x< 4\)
\(=>0\le x< 4\)
Bài 5:
- Áp dụng bất đẳng thức Cô-si ta có:
\(\sqrt{ac}\le\dfrac{a+c}{2}\Leftrightarrow a\sqrt{ac}\le\dfrac{a^2+ca}{2}\left(1\right)\)
- Chứng minh tương tự, ta có:
\(b\sqrt{ba}\le\dfrac{b^2+ab}{2}\left(2\right)\) ; \(c\sqrt{cb}\le\dfrac{c^2+bc}{2}\left(3\right)\)
- Từ \(\left(1\right),\left(2\right),\left(3\right)\) suy ra:
\(a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}\le\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\left(4\right)\)
Ta có: \(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}=2a^2\left(5\right)\)
- Chứng minh tương tự, ta có:
\(\dfrac{b^3}{c}+bc\ge2b^2\left(6\right)\) ; \(\dfrac{c^3}{a}+ca\ge2c^2\left(7\right)\)
- Từ \(\left(5\right),\left(6\right),\left(7\right)\) suy ra:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2a^2+2b^2+2c^2=\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}+\dfrac{3\left(a^2+b^2+c^2-ab-bc-ca\right)}{2}\)
Mặt khác ta bất đẳng thức phụ:
\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}+\dfrac{3.0}{2}=\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\left(8\right)\)
- Từ \(\left(4\right),\left(8\right)\Rightarrowđpcm\)
Dấu "=" xảy ra khi \(a=b=c\)
