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Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).
Áp dụng bđt Schwars ta có:
\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).
Đẳng thức xảy ra khi a = b = c = 1.
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Bài 1:
a: \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=2+2\sqrt{5}+2-2\sqrt{5}=4\)
Khi x=4 thì \(P=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)
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a, ĐKXĐ: \(x\ge0;x\ne9\)
b, rút gọn
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x}{x-9}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{x-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{x+1}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
c,Cho \(A\le-\dfrac{1}{3}\)
\(< =>\dfrac{3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ < =>\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}\le0\\ < =>\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\dfrac{\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\sqrt{x}-6\le0\\ < =>\sqrt{x}\le36\\ < =>0\le x\le36\)
Vậy để \(A\le-\dfrac{1}{3}\) thì \(0\le x\le36\)và\(x\ne9\)
d, \(A=\dfrac{-3}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}+3\ge3\\ =>\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\\ =>\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}\\ =-1\)
Vậy GTNN của A=-1
Xấu ''='' xảy ra khi \(\sqrt{x}=0\\ \Leftrightarrow x=0\)
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1.ĐK:\(x\ge0,x\ne9\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}-3}\)
\(=\left[\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}-5}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-5\right)}.\)
Để \(P< \dfrac{-1}{2}\Leftrightarrow\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-5\right)}< \dfrac{-1}{2}\)
Lời giải:
Áp dụng BĐT Cô-si:
$t(3-t)\leq \left(\frac{t+3-t}{2}\right)^2=\frac{9}{4}$
$\Rightarrow A\geq \frac{4(4t^2+9)}{9t}$
$=\frac{16t^2+36}{9t}=\frac{16t}{9}+\frac{4}{t}$
$\geq 2\sqrt{\frac{16t}{9}.\frac{4}{t}}=\frac{16}{3}$ (tiếp tục áp dụng BĐT Cô-si)
Vậy $A_{\min}=\frac{16}{3}$. Giá trị này đạt được khi $x=\frac{3}{2}$