Tìm GTNN của hàm số \(f\left(x\right)=\dfrac{4}{x}+\dfrac{9}{1-x}\) với \(0< x< 1\)
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c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
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1.
\(f\left(x\right)=\dfrac{4}{x}+\dfrac{x-1+1}{1-x}=\dfrac{2^2}{x}+\dfrac{1}{1-x}-1\ge\dfrac{\left(2+1\right)^2}{x+1-x}-1=8\)
\(f\left(x\right)_{min}=8\) khi \(x=\dfrac{2}{3}\)
2.
\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{1-x}\ge\dfrac{4}{x+1-x}=4\)
\(f\left(x\right)_{min}=4\) khi \(x=\dfrac{1}{2}\)
f(x)=4x+x−1+11−x=22x+11−x−1≥(2+1)2x+1−x−1=8f(x)=4x+x−1+11−x=22x+11−x−1≥(2+1)2x+1−x−1=8
f(x)min=8f(x)min=8 khi x=23x=23
2.
f(x)=1x+11−x≥4x+1−x=4f(x)=1x+11−x≥4x+1−x=4
f(x)min=4f(x)min=4 khi x=12
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a) \(f(x)\geq 2\sqrt{x^2.\frac{16}{x^2}}=2\sqrt{16}=2.4=8\)
Dấu "=" xảy ra khi và chỉ khi \(x^2=\frac{16}{x^2}\)
\(\Leftrightarrow x=2\)
Vậy GTNN của \(f(x)\) bằng 8 khi x=2
b) \(f(x)=\frac{1-x+x}{x}+\frac{2-2x+2x}{1-x}\)
\(f(x)=\frac{1-x}{x}+\frac{2x}{1-x}+3\)
\(f(x)\geq 2\sqrt{\frac{1-x}{x}.\frac{2x}{1-x}}+3=2\sqrt{2}+3\)
Dấu "=" xảy ra khi và chỉ khi \(\frac{1-x}{x}=\frac{2x}{1-x}\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTNN của \(f(x)\) bằng \(2\sqrt{2} +3\) khi \(x=\frac{1}{2}\)
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\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
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a, \(y=\dfrac{\sqrt{x-2}}{x}=\sqrt{\dfrac{1}{x}-\dfrac{2}{x^2}}\ge0\)
\(min=0\Leftrightarrow\dfrac{1}{x}-\dfrac{2}{x^2}=0\Leftrightarrow x=2\)
b, Áp dụng BĐT Cosi:
\(f\left(x\right)=\dfrac{x}{\sqrt{x-1}}=\dfrac{x-1+1}{\sqrt{x-1}}=\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\ge2\)
\(minf\left(x\right)=2\Leftrightarrow x=2\)
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a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
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\(f\left(1-x\right)+f\left(x\right)=\dfrac{9^{1-x}}{9^{1-x}+3}+\dfrac{9^x}{9^x+3}=\dfrac{9}{9+3.9^x}+\dfrac{9^x}{9^x+3}=\dfrac{3}{9^x+3}+\dfrac{9^x}{9^x+3}=1\)
\(\Rightarrow f\left(x\right)=1-f\left(1-x\right)\)
\(\Rightarrow f\left(cos^2x\right)=1-f\left(sin^2x\right)\)
Do đó:
\(f\left(3m+\dfrac{1}{4}sinx\right)+f\left(cos^2x\right)=1\)
\(\Leftrightarrow f\left(3m+\dfrac{1}{4}sinx\right)=f\left(sin^2x\right)\) (1)
Hàm \(f\left(x\right)=\dfrac{9^x}{9^x+3}\) có \(f'\left(x\right)=\dfrac{3.9^x.ln9}{\left(9^x+3\right)^2}>0\Rightarrow f\left(x\right)\) đồng biến trên R
\(\Rightarrow\left(1\right)\Leftrightarrow3m+\dfrac{1}{4}sinx=sin^2x\)
Đến đây chắc dễ rồi, biện luận để pt \(sin^2x-\dfrac{1}{4}sinx=3m\) có 8 nghiệm trên khoảng đã cho
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\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
Áp dụng bất đẳng thức Svacxo, ta có:
\(f\left(x\right)=\dfrac{4}{x}+\dfrac{9}{1-x}=\dfrac{2^2}{x}+\dfrac{3^2}{1-x}\ge\dfrac{\left(2+3\right)^2}{x+1-x}=25\)
Vậy \(f\left(x\right)_{min}=25\Leftrightarrow\dfrac{2}{x}=\dfrac{3}{1-x}\Leftrightarrow x=\dfrac{2}{5}\)