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\(2a^2+\frac{1}{a^2}+\frac{b^2}{4}=4\Leftrightarrow\left(a^2+\frac{1}{a^2}-2\right)+\left(a^2+\frac{b^2}{4}-ab\right)=4-ab-2\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(a-\frac{b}{2}\right)^2=2-ab\)
\(VF=2-ab=\left(a-\frac{1}{a}\right)^2+\left(b-\frac{b}{2}\right)^2\ge0\)
Hay \(ab\le2\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=\frac{1}{a}\\b=\frac{b}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(a;b\right)=\left(1;\frac{1}{2}\right)\\\left(a;b\right)=\left(-1;-\frac{1}{2}\right)\end{cases}}\)
Để H lớn nhất thì \(\frac{1}{H}=\frac{\left(x+2018\right)^2}{x}\) nhỏ nhất.
Ta có: \(\frac{1}{H}=\frac{x^2+2.x.2018+2018^2}{x}=x+4036+\frac{2018^2}{x}\)
\(\frac{x+\frac{2018^2}{x}}{2}\ge\sqrt{x.\frac{2018^2}{x}}=2018\) (áp dụng bất đẳng thức cosi) \(\Rightarrow x+\frac{2018^2}{x}\ge4036\)
\(\frac{1}{A}\ge4036+4036=8072\Rightarrow A\le\frac{1}{8072}\)
Dấu "=" xảy ra khi: \(x=\frac{2018^2}{x}\Rightarrow x^2=2018^2\Rightarrow x=2018\left(x>0\right)\)
Vậy GTLN của H là \(\frac{1}{8072}\Leftrightarrow x=2018\)
\(A=\dfrac{x^2+mx+n}{x^2+2x+4}\)
\(\Leftrightarrow Ax^2+2Ax+4A=x^2+mx+n\)
\(\Leftrightarrow\left(A-1\right)x^2+\left(2A-m\right)x+\left(4A-n\right)=0\left(1\right)\)
A có cực trị khi (1) có nghiệm
\(\Leftrightarrow\Delta=\left(4A^2-4Am+m^2\right)-4\left[4A^2-A\left(n+4\right)+n\right]\ge0\)
\(\Leftrightarrow-12A^2-4A\left(m-n-4\right)+m^2-4n\ge0\) (1)
Mặt khác, theo gt, ta có: \(\left\{{}\begin{matrix}A\ge\dfrac{1}{3}\\A\le3\end{matrix}\right.\)
\(\Rightarrow\left(3A-1\right)\left(3-A\right)\ge0\)
\(\Leftrightarrow-3A^2+10A-3\ge0\)
\(\Leftrightarrow-12A^2+40A-12\ge0\) (2)
Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}m-n-4=-10\\m^2-4n=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+6=n\\m^2-4\left(m+6\right)=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=m+6\\\left(m-6\right)\left(m+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}n=12\\n=4\end{matrix}\right.\\\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(m;n\right)=\left(6;12\right);\left(-2;4\right)\)
Lời giải:
Áp dụng BĐT Cô-si cho các số dương ta có:
\(4=2a^2+\frac{b^2}{4}+\frac{1}{a^2}=a^2+a^2+\frac{b^2}{4}+\frac{1}{a^2}\geq 4\sqrt[4]{a^2.a^2.\frac{b^2}{4}.\frac{1}{a^2}}\)
\(\Rightarrow 1\geq \frac{a^2b^2}{4}\Rightarrow a^2b^2\leq 4\Rightarrow -2\leq ab\leq 2\)
Do đó:
\(-2+2019\leq ab+2019\leq 2+2019\Leftrightarrow 2017\leq S\leq 2021\)
Vậy \(S_{\min}=2017\Leftrightarrow (a,b)=(1;-2)\) hoặc \((-1;2)\)
\(S_{\max}=2021\Leftrightarrow (a,b)=(1;2)\) hoặc \((-1;-2)\)
\(P=-\dfrac{2019}{x^2}+\dfrac{m}{x}=-2019\left(\dfrac{1}{x^2}-2.\dfrac{m}{2.2019}.\dfrac{1}{x}\right)\)
\(=-2019\left(\dfrac{1}{x^2}-2.\dfrac{m}{4038}.\dfrac{1}{x}+\dfrac{m^2}{4038^2}-\dfrac{m^2}{4038^2}\right)=-2019\left(\dfrac{1}{x}-\dfrac{m}{4038}\right)^2+\dfrac{2019m^2}{4038^2}\le\dfrac{2019m^2}{4038^2}\)
\(\Rightarrow\dfrac{2019m^2}{4038^2}=2019\Rightarrow m=\pm4038\)
\(P=\dfrac{mx-2019}{x^2}\Rightarrow px^2-mx+2019=0\)
\(\Delta=m^2-4.2019P\ge0\)
\(\Leftrightarrow P\le\dfrac{m^x}{8076}\)
để \(\max\limits_P=2019\) thì \(\dfrac{m^2}{8076}=2019\)
\(\Leftrightarrow m^2=8076.2019\)
\(=2.2.2019.2019\)
\(\Leftrightarrow m=4038\)(vì m>0)
vậy m=4038