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2.
a/ Áp dụgn hệ quả bđt cô si,ta có :
\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)
Vậy GTLN A =a^2/3 khi x= y =z =a/3
b/Áp dụng BĐT Cô-Si dạng Engel,ta có :
\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)
Vậy GTNN của B = a^2/2 khi x=y=z =a/3
\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
1.Ta có : \(A=\dfrac{2}{\sqrt{x-3}}+\dfrac{1}{\sqrt{x+3}}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow M=A\div B=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\times\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x+3}}\)
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
Lời giải:
\(S=a^2+\frac{18}{\sqrt{a}}=a^2(1-\frac{1}{2\sqrt{6}})+\frac{a^2}{2\sqrt{6}}+\frac{18}{\sqrt{a}}\)
Áp dụng BĐT AM-GM:
\(\frac{a^2}{2\sqrt{6}}+\frac{18}{\sqrt{a}}\geq 2\sqrt{\frac{3\sqrt{6}.\sqrt{a^3}}{2}}\geq 2\sqrt{\frac{3\sqrt{6}.\sqrt{6^3}}{2}}=6\sqrt{6}\) (do $a\geq 6$)
\(a^2(1-\frac{1}{2\sqrt{6}}\geq 6^2(1-\frac{1}{2\sqrt{6}})=36-3\sqrt{6}\) (do $a\geq 6$)
Cộng lại:
\(\Rightarrow S\ge 36+3\sqrt{6}\)
Vậy $S_{\min}=36+3\sqrt{6}$ khi $a=6$