Lời giải:
Đặt \(\sqrt{x}=a(a\ge 0)\)
Khi đó: \(P=\frac{4a}{3(a^2-a+1)}\)
Để \(P=\frac{8}{9}\Rightarrow \frac{4a}{3(a^2-a+1)}=\frac{8}{9}\)
\(\Rightarrow \frac{a}{a^2-a+1}=\frac{2}{3}\Rightarrow 3a=2(a^2-a+1)\)
\(\Leftrightarrow 2a^2-5a+2=0\Leftrightarrow (a-2)(2a-1)=0\)
\(\Rightarrow \left[\begin{matrix} a-2=0\\ 2a-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=2=\sqrt{x}\\ a=\frac{1}{2}=\sqrt{x}\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x=4\\ x=\frac{1}{4}\end{matrix}\right.\) (t/m)
b)
Vì \(a\geq 0; a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}>0\)
Do đó: \(P=\frac{4}{3}.\frac{a}{a^2-a+1}\geq \frac{4}{3}.0=0\)
Vậy \(P_{\min}=0\Leftrightarrow a=0\Leftrightarrow x=0\)
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Áp dụng BĐT Cô-si: \(a^2+1\geq 2a\Rightarrow a^2-a+1\geq 2a-a=a\)
\(\Rightarrow \frac{a}{a^2-a+1}\leq \frac{a}{a}=1\Rightarrow P=\frac{4}{3}.\frac{a}{a^2-a+1}\leq \frac{4}{3}.1=\frac{4}{3}\)
Vậy \(P_{\max}=\frac{4}{3}\Leftrightarrow a=1\Leftrightarrow x=1\)
Bài 2:
Đặt \(P=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-\sqrt{12-4\sqrt{5}}\)
Có:
\(4+\sqrt{15}=\frac{8+2\sqrt{15}}{2}=\frac{5+3+2\sqrt{3.5}}{2}=\frac{(\sqrt{3}+\sqrt{5})^2}{2}\)
\(\Rightarrow \sqrt{4+\sqrt{15}}=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{2}}\)
Tương tự: \(\sqrt{4-\sqrt{15}}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(12-4\sqrt{5}=12-2\sqrt{20}=10+2-2\sqrt{10.2}=(\sqrt{10}-\sqrt{2})^2\)
\(\Rightarrow \sqrt{12-4\sqrt{5}}=\sqrt{10}-\sqrt{2}\)
Vậy \(P=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{2}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-(\sqrt{10}-\sqrt{2})\)
\(=\sqrt{2}\)
