TÌM GTLN của biểu thức \(P=\frac{\sqrt{a-2}}{a},với\)\(a\ge2\)
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a) ĐK : \(x\ge0\)
A = \(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{1}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\cdot\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
b) \(A=\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{x-\sqrt{x}+1-x+2\sqrt{x}-1}{x-\sqrt{x}+1}=1-\frac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le1\)
=> Max A = 1
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\)<=> x = 1
Vậy Max A = 1 <=> x = 1
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*Sửa đề: tìm GTNN
\(A=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\)
\(=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\ge\frac{\frac{2+c-2}{2}}{\sqrt{2}c}=\frac{\frac{c}{2}}{\sqrt{2}c}=\frac{1}{2\sqrt{2}}\)
TƯơng tự cho 2 BĐT còn lại ta cũng có:
\(\frac{\sqrt{a-3}}{a}\ge\frac{1}{2\sqrt{3}};\frac{\sqrt{b-4}}{b}\ge\frac{1}{2\sqrt{4}}\)
Suy ra \(A\ge\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)\)
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a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
=> \(A=\left(\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{x\sqrt{x}-1}\right)\left(\frac{2}{\sqrt{x}-1}\right)\)
=> \(A=\left(\frac{x-2\sqrt{x}+1}{x\sqrt{x}-1}\right)\left(\frac{2}{\sqrt{x}-1}\right)\)
=> \(A=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{2}{x+\sqrt{x}+1}\)
c, Ta có : \(A=\frac{2}{x+\sqrt{x}+1}=\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}\)
Ta thấy \(\frac{2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}}>0\forall x\ne1\)
Ta có : \(\frac{1}{P}=\frac{a}{\sqrt{a-2}}=\frac{a-2+2}{\sqrt{a-2}}=\sqrt{a-2}+\frac{2}{\sqrt{a-2}}\)
Áp dụng BĐT Cauchy :
\(\sqrt{a-2}+\frac{2}{\sqrt{a-2}}\ge2\sqrt{\sqrt{a-2}.\frac{2}{\sqrt{a-2}}}=2\sqrt{2}\)
\(\Leftrightarrow P\ge\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{a-2}=\frac{2}{\sqrt{a-2}}\Leftrightarrow a-2=2\Leftrightarrow a=4\)
Vậy maxP =\(\frac{\sqrt{2}}{4}\)<=> a = 4