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$D=x+\dfrac{4}{x-1}$
$=x-1+\dfrac{4}{x-1}+1$
Áp dụng bất đẳng thức Cauchy ( Cô-si) có:
$(x-1)+\dfrac{4}{x-1}≥2.\sqrt[]{(x-1).\dfrac{4}{x-1}}=4$
$⇒(x-1)+\dfrac{4}{x-1}+1≥5$
Hay $D≥5$
Dấu $=$ xảy ra $⇔x-1=\dfrac{4}{x-1}⇔(x-1)^2=4⇔x=3 (x≥3)$
$E=\dfrac{x^2+3}{x+1}$
$=\dfrac{x^2-1+4}{x+1}$
$=x-1+\dfrac{4}{x+1}$
$=\dfrac{x+1}{25}+\dfrac{4}{x+1}+\dfrac{24.(x+1)}{25}-1$
Áp dụng bất đẳng thức Cô-si cóL
$\dfrac{x+1}{25}+\dfrac{4}{x+1}≥2.\sqrt[]{\dfrac{x+1}{25}.\dfrac{4}{x+1}}=\dfrac{4}{5}$
$x≥9⇒\dfrac{24.(x+1)}{25}≥\dfrac{24.10}{25}=\dfrac{48}{5}$
$⇒E≥\dfrac{4}{5}+\dfrac{48}{5}=\dfrac{52}{5}$
Dấu $=$ xảy ra $⇔x=9$
$F=\dfrac{x^2-5x+7}{x-1}$
$=\dfrac{x^2-2x+1-3x+3+4}{x-1}$
$=\dfrac{(x-1)^2-3.(x-1)+4}{x-1}$
$=x-1-3+\dfrac{4}{x-1}$
$=\dfrac{x-1}{3}+\dfrac{3}{x-1}+\dfrac{x-1}{9}+\dfrac{1}{x-1}+\dfrac{5.(x-1)}{9}$
$≥2.\sqrt[]{\dfrac{x-1}{3}.\dfrac{3}{x-1}}+2.\sqrt[]{\dfrac{x-1}{9}.\dfrac{1}{x-1}}+\dfrac{5.(4-1)}{9}$
$≥2+\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{13}{3}$
Dấu $=$ xảy ra $⇔x=4$
