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Áp dụng BĐT Cô si ta có:
\(x^3+8y^3+1\ge3\sqrt[3]{x^3\cdot8y^3\cdot1}=6xy\)
\(\Rightarrow x^3+8y^3+1-6xy\ge0\)
Dấu "=" xảy ra tại \(x=2y=1\Rightarrow x=1;y=\frac{1}{2}\)
Khi đó:
\(A=x^{2018}+\left(y-\frac{1}{2}\right)^{2019}=1^{2018}+0^{2019}=1\)
\(2x^2+2y^2+z^2-2x+2y+2xy+2yz+2zx+2=0\)
\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\)\(\left(x+y\right)^2+\left(y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\)\(x=-y=z=1\)
\(\Rightarrow\)\(A=x^{2018}+y^{2018}+z^{2018}=1^{2018}+\left(-1\right)^{2018}+1^{2018}=3\)
...
\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)
\(\Leftrightarrow x^{2018}=3^{2018}\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)
Đến đây tự tính A nha!
P=2x+y+30x+5y
=(6x5+30x)+(y5+5y)+(4x5+4y5)
≥2.6+2+45.10=22
Vậy GTNN là P = 22 khi x = y = 5
Ta có: \(P=\frac{9x+18y}{xy}+\frac{2x-5y}{12}+2018\)
\(=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2018\)
\(=\frac{18}{x}+\frac{x}{2}+\frac{9}{y}+\frac{y}{4}-\frac{x}{3}-\frac{2y}{3}+2018\)
\(=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x+2y}{3}+2018\)
Vì \(x,y>0\Rightarrow\frac{18}{x}>;\frac{x}{2}>0\)
Áp dụng BĐT cô si cho hai số dương ta có:
\(\frac{18}{x}+\frac{x}{2}\ge2\sqrt{\frac{18}{x}.\frac{x}{2}}=6\)
\(\frac{9}{y}+\frac{y}{4}\ge2\sqrt{\frac{9}{y}.\frac{y}{4}}=3\)
Vì \(x+2y\le18\)
\(\Rightarrow\frac{x+2y}{3}\le\frac{18}{3}=6\)
\(\Rightarrow\frac{-x+2y}{3}\ge-6\)
\(\Rightarrow P\ge6+3-6+2018\)
\(\Rightarrow P\ge2021\)
\(\Rightarrow MinP=2021\Leftrightarrow\hept{\begin{cases}\frac{18}{x}=\frac{x}{2}\\\frac{9}{y}=\frac{y}{4}\\x+2y=18\end{cases}}\)và x,y>0
\(\Leftrightarrow\hept{\begin{cases}x=6\\y=6\end{cases}\Rightarrow x=y=6}\)
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