Ta có: 3x + y = 1 => y = 1 - 3x

=> M = 3x² + y² = 3x² + (1-3x)² 

         = 3x² + 1 - 6x + 9x² 

         = 12x² - 6x + 1

         = 12.(x² -\(\frac{1}{2}x\) + \(\frac{1}{12}\))

         = 12.((x² - 2. \(\frac{1}{4}x\)+ \(\frac{1}{16}\)) - \(\frac{1}{16}\)+ \(\frac{1}{12}\))

         = 12.((x-\(\frac{1}{4}\))²  + \(\frac{1}{48}\)⁾

           ^{= 12.} (x-\(\frac{1}{4}\))² + \(\frac{1}{4}\)     

^{=> M     \(\ge\)}\(\frac{1}{4}\)

Dấu ''='' xảy ra khi: (x - \(\frac{1}{4}\))² = 0 => x = \(\frac{1}{4}\)

Vậy Mmin= \(\frac{1}{4}\)khi x= \(\frac{1}{4}\)

Ta có: \(3x+y-1=0\)

\(\Rightarrow3x+y=1\)

Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có: 

 \(\left(3x^2+y^2\right)\left(3+1\right)=\left[\left(\sqrt{3}x\right)^2+y^2\right]\left[\left(\sqrt{3}\right)^2+1^2\right]\ge\left(\sqrt{3}x.\sqrt{3}+y.1\right)^2\)

\(\Leftrightarrow4B\ge1^2\)

\(\Leftrightarrow B\ge\frac{1}{4}\)

Dấu = xảy ra khi \(\frac{\sqrt{3}x}{\sqrt{3}}=\frac{y}{1}\Rightarrow x=y=\frac{1}{4}\)

Vậy........

Từ đk trên ta có:  \(2y^2+2zy+2z^2=2-3x^2\)

<=> \(3x^2+2y^2+2zy+2z^2=2\left(1\right)\)

<=>\(\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

Do (x-y)²≥0; (x-z)²≥0 nên từ(*) suy ra (x+y+z)²≤2

Hay \(-\sqrt{2}\le x+y+z\le\sqrt{2}\)

Dấu "=" xảy ra khi x-y =0 và x-z=0 hay x=y=z

Thay vào (1) ta được 9x²=2 ; x=\(\dfrac{\sqrt{2}}{3};\dfrac{-\sqrt{2}}{3}\)

Với x=y=z =x=\(\dfrac{\sqrt{2}}{3};\dfrac{-\sqrt{2}}{3}\)thì max=\(\sqrt{2}\), min =\(-\sqrt{2}\)

Ta có : \(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

\(\Rightarrow-\sqrt{2}\le B\le\sqrt{2}\)

Vậy \(MinB=-\sqrt{2}\Leftrightarrow x=y=z=-\frac{\sqrt{2}}{3}\)

\(MaxB=\sqrt{2}\Leftrightarrow x=y=z=\frac{\sqrt{2}}{3}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$P^2\leq (x+y)[(29x+3y)+(29y+3x)]=32(x+y)^2\leq 32.(x^2+y^2)(1+1)=64(x^2+y^2)\leq 64.2=128$

$\Rightarrow P\leq 8\sqrt{2}$
Vậy $P_{\max}=8\sqrt{2}$