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\(3x^2+6y^2+2z^2+3y^2z^2-18x=6\)
\(\Leftrightarrow3\left(x-3\right)^2+6y^2+2z^2+3y^2z^2=33\)
\(\Rightarrow3\left(x-3\right)^2\le33\)
\(\Leftrightarrow\left(x-3\right)^2\le11\)
\(\Leftrightarrow\left(x-3\right)^2=\left\{0;1;4;9\right\}\)
Thế lần lược vô giải tiếp sẽ ra
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
Từ giả thiết \(xy+yz+zx=5\)
ta có \(x^2+5=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\)
Áp dụng BĐT AM-GM , ta có
\(\sqrt{6\left(x^2+5\right)}=\sqrt{6\left(x+y\right)\left(z+x\right)}\le\frac{3\left(x+y\right)+2\left(z+x\right)}{2}=\frac{5x+3y+2z}{2}\)
CM tương tự ta được \(\sqrt{6\left(y^2+5\right)}\le\frac{3x+5y+2z}{2};\sqrt{z^2+5}\le\frac{x+y+2z}{2}\)
Cộng zế zới zế BĐt trên ta đc
\(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}\le\frac{9x+9y+6z}{2}\)
\(=>P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{x^2+5}}\ge\frac{2\left(3x+3y+2z\right)}{9x+9y+6z}=\frac{2}{3}\)
=> \(GTNN\left(P\right)=\frac{2}{3}khi\left(x=y=1;z=2\right)\)
Ta có \(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}=\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}\)\(+\sqrt{6\left(z+x\right)\left(z+y\right)}\)
\(\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(x+y\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}\le\frac{9x+9y+6z}{2}=\frac{3}{2}\)\(\left(3x+3y+2z\right)\)
\(\Rightarrow P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\ge\frac{2}{3}\)
dấu "=" xảy ra \(\Leftrightarrow x=y=1;z=2\)
Vậy \(P_{min}=\frac{2}{3}\Leftrightarrow x=y=1;z=2\)
\(\hept{\begin{cases}x^3-3x-2=2-y\\y^3-3y-2=4-2z\\z^3-3z-2=6-3x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^3-x-2x-2=2-y\\y^3-y-2y-2=2\left(2-z\right)\\z^3-z-2z-2=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\left(x^2-1\right)-2\left(x+1\right)=2-y\\y\left(y^2-1\right)-2\left(y+1\right)=2\left(2-z\right)\\z\left(z^2-1\right)-2\left(z+1\right)=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left[x\left(x-1\right)-2\right]=2-y\\\left(y+1\right)\left[y\left(y-1\right)-2\right]=2\left(2-z\right)\\\left(z+1\right)\left[z\left(z-1\right)-2\right]=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(x^2-x-2\right)=2-y\\\left(y+1\right)\left(y^2-y-2\right)=2\left(2-z\right)\\\left(z+1\right)\left(z^2-z-2\right)=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\left(x-2\right)=2-y\\\left(y+1\right)^2\left(y-2\right)=2\left(2-z\right)\\\left(z+1\right)^2\left(z-2\right)=3\left(2-x\right)\end{cases}}\)
Nhân các vế của 3 phương trình với nhau ta được:
\(\left(x+1\right)^2\left(x-2\right)\left(y+1\right)^2\left(y-2\right)\left(z+1\right)^2\left(z-2\right)=6\left(2-y\right)\left(2-z\right)\left(2-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=-6\left(y-2\right)\left(z-2\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\left(y-2\right)\left(x-2\right)\left(z-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left[\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\right]=0\)
Vì \(\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6>0\)
Nên \(\left(x-2\right)\left(y-2\right)\left(z-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y-2=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\\z=2\end{cases}}}\)
Vậy x = y = z = 2
<=>3(x2-6x+9)+6y2+2z2+3y2z2=33
<=>3(x-3)2+6y2+2z2+3y2z2=33
nhận thấy 3(x-3)2;6y2;3y2z2 chia hết cho
=>2z2 chia hết cho 3=>z chia hết cho 3
giả sử trong 4 số đó không số nào =0
=>\(3\left(x-3\right)^2\ge3;6y^2\ge6;2z^2\ge18;3y^2z^2\ge27\Rightarrow3\left(x-3\right)^2+6y^2+2z^2+3y^2z^2\ge54\)(vô lí)
với x-3=0
=>x=3
pt trở thành 6y2+2z2+3y2z2=6
<=>(3y2+2)(z2+2)=10
với y=0
=>3(x-3)2+2z2=33 (đến đây thid dễ rồi)
với z=0=>3(x-3)2+6y2=33
=>(x-3)2+2y2=11