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\(P=\dfrac{x^2+y^2+6}{x+y}=\dfrac{x^2+y^2+2xy+4}{x+y}=\dfrac{\left(x+y\right)^2+4}{x+y}=x+y+\dfrac{4}{x+y}\)
\(P\ge2\sqrt{\left(x+y\right).\dfrac{4}{x+y}}=4\)
\(P_{min}=4\) khi \(x=y=1\)
\(K=\left(4xy+\dfrac{1}{4xy}\right)+\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{5}{4xy}\)
\(K\ge2\sqrt{\dfrac{4xy}{4xy}}+\dfrac{4}{x^2+y^2+2xy}+\dfrac{5}{\left(x+y\right)^2}\ge2+4+5=11\)
\(K_{min}=11\) khi \(x=y=\dfrac{1}{2}\)
\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)
Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)
\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)
\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)
\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)
\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)
\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)
x+y=1=>y=1-x
\(Q=2x^2-y^2+x+\frac{1}{x}+2020\)\(=2x^2-\left(1-x\right)^2+x+\frac{1}{x}+2020\)\(=2x^2-\left(1-2x+x^2\right)+x+\frac{1}{x}+2020\)\(=2x^2-1+2x-x^2+x+\frac{1}{x}+2020\)
\(=\left(x^2+2x+1\right)+\left(x+\frac{1}{x}\right)+2018\)\(=\left(x+1\right)^2+\left(x+\frac{1}{x}\right)+2018\)
Ta có: \(\left(x+1\right)^2\ge0\forall x>0\)
Áp dụng BĐT Cô-si cho 2 số dương \(x\)và \(\frac{1}{x}\):
\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)
\(\Rightarrow Q\ge2+2018=2020\)
Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+1=0\\x=\frac{1}{x}\end{cases}\Leftrightarrow x=-1}\)\(\Rightarrow y=1-\left(-1\right)=2\)
Vậy \(minQ=2020\Leftrightarrow x=-1;y=2\)
Lời giải:
Áp dụng BĐT Cô-si và Cauchy-Schwarz cho các số dương ta có:
$A=\frac{1}{x}+\frac{1}{\sqrt{xy}}\geq \frac{1}{x}+\frac{1}{\frac{x+y}{2}}=\frac{1}{x}+\frac{2}{x+y}=2(\frac{1}{2x}+\frac{1}{x+y})$
$\geq 2.\frac{4}{2x+x+y}=\frac{8}{3x+y}\geq \frac{8}{4}=2$
Vậy $A_{\min}=2$. Giá trị này đạt được tại $x=y; 3x+y=4\Leftrightarrow x=y=1$
