G = \(\dfrac{x^2}{x-1}\)
= \(\dfrac{x^2-4x+4+4x-4}{x-1}\)
= \(\dfrac{\left(x-2\right)^2+4\left(x-1\right)}{x-1}\)
= \(\dfrac{\left(x-2\right)^2}{x-1}+4\)
Vì x>1 nên \(\left\{{}\begin{matrix}\left(x-2\right)^2\text{≥}0\\x-1>0\end{matrix}\right.\)
=> G ≥ 4
=> G = 4 đạt GTNN
Dấu bằng xảy ra <=> \(\left(x-2\right)^2=0\)
<=> \(x=2\)
\(Do\) \(x>2\)
\(=>\left\{{}\begin{matrix}x-2\text{ ≥0}\\2x-1>0\end{matrix}\right.\)
\(=>\left(x-2\right)\left(2x-1\right)\text{ ≥0}\)
\(< =>2x^2-5x+2\text{≥}0\)
\(< =>2x^2+2\text{≥}5x\)
\(< =>2x+\dfrac{2}{x}\text{≥}5\)
\(< =>x+\dfrac{1}{x}\text{≥}2,5\)
\(< =>H\text{≥}2,5\)
\(< =>H=5\) \(đạt\) \(GTNN\)
Dấu bằng xảy ra khi \(x-2=0< =>x=2\)
\(K=x^2+\dfrac{1}{x}\)
\(=\dfrac{53x^3}{54}+\left(\dfrac{x^2}{54}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)\)
Áp dụng BĐT Cô si cho 3 số dương
\(\dfrac{x^2}{54}+\dfrac{1}{2x}+\dfrac{1}{2x}\text{≥}3.\sqrt[3]{\dfrac{x^2}{54}.\dfrac{1}{2x}.\dfrac{1}{2x}}\)\(\text{≥}\dfrac{53.9}{54}+3.\sqrt[3]{54.4}\)\(=\dfrac{28}{3}\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{x^2}{54}=\dfrac{1}{2x}=\dfrac{1}{2x}\\x=3\end{matrix}\right.\)\(< =>x=3\)
\(H=x+\dfrac{1}{x}\)
\(=\dfrac{3}{4}x+\left(\dfrac{1}{4}x+\dfrac{1}{x}\right)\)
Áp dụng BĐT Cô si cho 2 số dương \(\dfrac{1}{4}x\) và \(\dfrac{1}{x}\)
\(\dfrac{1}{4}x+\dfrac{1}{x}\text{≥}2.\sqrt{\dfrac{1}{4}x.\dfrac{1}{x}}\)\(=1\)
\(< =>H\text{≥}\dfrac{3}{4}.2+1=\dfrac{5}{2}\)
\(< =>H=\dfrac{5}{2}đạt\) \(GTNN\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{4}x=\dfrac{1}{x}\\x=2\end{matrix}\right.< =>x=2\)