Ai giúp mik vs

Huhu ai giúp vs

Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)

<=>\(x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+yz+zx\right)\)<=>\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

<=>\(3^2\ge3\left(xy+yz+zx\right)\)<=>\(P=xy+yz+zx\le3\)=>P_max=3 <=> x=y=z=1

Ta có BĐT đúng sau:

x2 + y2 + z2 >= xy + yz + zx

<=> (x + y + z)2 >= 3(xy + yz + zx)

<=> 9 >= 3 P <=> P <=3 (dấu bằng khi x = y = z =1)

\(A=3yz+\left(4-y-z\right)\left(y+2z\right)\)

\(A=-y^2+4y-2z^2+8z\)

\(A=-\left(y-2\right)^2-2\left(z-2\right)^2+12\le12\)

\(A_{max}=12\) khi \(\left(x;y;z\right)=\left(0;2;2\right)\)

Ta có: \(B=-x^2-y^2-xy+2x+3y\)

\(=-\frac14\left(4x^2+4y^2+4xy-8x-12y\right)\)

\(=-\frac14\left\lbrack4x^2+4xy+y^2-8x-4y+3y^2-8y\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x+y\right)^2-4\left(2x+y\right)+4+3y^2-8y-4\right\rbrack\)

\(=\frac{-1}{4}\left\lbrack\left(2x+y-2\right)^2+3\left(y^2-\frac83y-\frac43\right)\right\rbrack\)

\(=\frac{-1}{4}\left\lbrack\left(2x+y-2\right)^2+3\left(y^2-2\cdot y\cdot\frac43+\frac{16}{9}-\frac{16}{9}-\frac43\right)\right\rbrack\)

\(=\frac{-1}{4}\left\lbrack\left(2x+y-2\right)^2+3\left(y^2-2\cdot y\cdot\frac43+\frac{16}{9}-\frac{28}{9}\right)\right\rbrack\)

\(=\frac{-1}{4}\left\lbrack\left(2x+y-2\right)^2+3\left(y-\frac43\right)^2-\frac{28}{3}\right\rbrack\le-\frac14\cdot\frac{-28}{3}=\frac73\forall x\)

Dấu '=' xảy ra khi \(\begin{cases}y-\frac43=0\\ 2x+y-2=0\end{cases}\Rightarrow\begin{cases}y=\frac43\\ 2x=-y+2=-\frac43+2=\frac23\end{cases}\Rightarrow\begin{cases}y=\frac43\\ x=\frac13\end{cases}\)

ko

bt

Xét \(x^2+y^2-xy=4\)

\(\Rightarrow x^2-2xy+y^2+xy=4\)

\(\Rightarrow\left(x-y\right)^2+xy=4\)

\(\Rightarrow xy=-\left(x-y\right)^2+4\)

Lại có: \(C=x^2+y^2=xy+4\)

\(=-\left(x-y\right)^2+4+4\)

\(=-\left(x-y\right)^2+8\)

Vì \(\left(x-y\right)^2\ge0\forall x,y\)

\(\Rightarrow-\left(x-y\right)^2\le0\forall x,y\)

\(\Rightarrow-\left(x-y\right)^2+8\le8\forall x,y\)

hay\(C\le8\forall x,y\)

GTLN là 8

Dấu "=" xảy ra khi: \(\left(x-y\right)^2=0\Rightarrow x=y\)

#DDN