1. Tìm GTNN m của hàm số f(x)= \(\dfrac{4}{x}\) + \(\dfrac{x}{1-x}\) với 1>x>0
2. Tìm GTNN m của hàm số f(x)= \(\dfrac{1}{x}\) + \(\dfrac{1}{1-x}\) với 0<x<1
Giúp mk với nhé thanks trước.
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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 \(x=\frac{1}{2}\)
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:
\(f\left(x\right)=\dfrac{x}{\sqrt{x-1}}=\dfrac{x-1+1}{\sqrt{x-1}}=\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\ge2\)
\(minf\left(x\right)=2\Leftrightarrow x=2\)
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 = \(\sqrt{6}\).
1: \(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2+2x-\left(m+1\right)=x^2+2x-m-1\)
\(\Delta=2^2-4\left(-m-1\right)=4m+8\)
Để f'(x)>=0 với mọi x thì 4m+8<=0 và 1>0
=>m<=-2
=>\(m\in\left\{-10;-9;...;-2\right\}\)
=>Có 9 số
\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\)
Áp dụng bđt Cô-si :
\(\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge3\sqrt[3]{\dfrac{1}{8}x\cdot\dfrac{1}{8}x\cdot\dfrac{8}{x^2}}=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge7+\dfrac{3}{2}=\dfrac{17}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow x=4\)
\(f\left(x\right)=\dfrac{x}{8}+\dfrac{x}{8}+\dfrac{8}{x^2}+\dfrac{7}{4}x\ge3\sqrt[3]{\dfrac{8x^2}{64x^2}}+\dfrac{7}{4}.4=\dfrac{17}{2}\)
Dấu "=" xảy ra khi \(x=4\)
2: ĐKXĐ: x<>1
\(f'\left(x\right)=\dfrac{\left(x^2-3x+3\right)'\left(x-1\right)-\left(x^2-3x+3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)
\(=\dfrac{\left(2x-3\right)\left(x-1\right)-\left(x^2-3x+3\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2-5x+3-x^2+3x-3}{\left(x-1\right)^2}=\dfrac{x^2-2x}{\left(x-1\right)^2}\)
f'(x)=0
=>x^2-2x=0
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
1:
\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)
=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)
f'(x)=0
=>\(\left(x-2\sqrt{2}\right)^2=0\)
=>\(x-2\sqrt{2}=0\)
=>\(x=2\sqrt{2}\)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a.
\(y=\dfrac{4}{x}+\dfrac{1}{1-x}-1\ge\dfrac{\left(2+1\right)^2}{x+1-x}-1=8\)
\(y_{min}=8\) khi \(x=\dfrac{4}{5}\)
b.
\(y=\dfrac{1}{x}+\dfrac{1}{1-x}\ge\dfrac{4}{x+1-x}=4\)
\(y_{min}=4\) khi \(x=\dfrac{1}{2}\)
\(y=\dfrac{x+3}{4}+\dfrac{9}{x-1}=\dfrac{x-1}{4}+\dfrac{9}{x-1}+1\)
\(y\ge2\sqrt{\dfrac{9\left(x-1\right)}{4\left(x-1\right)}}+1=4\)
\(y_{min}=4\) khi \(x=7\)
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