ai tích mình mình tích lại cho

ai ,mình tích  lại

2x^2+xy+2y^2 = 5/4.(x+y)^2 + 3/4. (x-y)^2 >= 5/4. (x+y)^2 
=> cbh(2x^2+xy+2y^2) >= cbh5 / 2. (x+y) 
tương tự với 2 căn còn lại.. cộng vế ta có VT >= cbh5 ( x+y+z) = cbh5 : dpcm 
dau = cay ra <=> x=y=z=1/3

a)\(P=x^2+4x+2xy+3y^2+5y+2017\)

\(=x^2+2xy+y^2+4y+4+4x+2y^2+y+\dfrac{1}{8}+\dfrac{16103}{8}\)

\(=\left(x+y+2\right)^2+2\left(y^2+\dfrac{y}{2}+\dfrac{1}{16}\right)+\dfrac{16103}{8}\)

\(=\left(x+y+2\right)^2+2\left(y+\dfrac{1}{4}\right)^2+\dfrac{16103}{8}\ge\dfrac{16103}{8}\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{7}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)

b)\(Q=-x^2+4x-3y^2+6y+2017\)

\(=-x^2+4x-4-3y^2+6y+3+2024\)

\(=-\left(x^2-4x+4\right)-\left(3y^2-6y-3\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y^2-2y-1\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\ge2024\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Ta có:

\(P=x^2+4x+2xy+3y^2+5y+2017\)

\(=x^2+2x\left(y+2\right)+\left(y+2\right)^2+2y^2+y+2013\)

\(=\left[x+\left(y+2\right)\right]^2+2\left(y^2+y+0,25\right)+2012,5\)

\(=\left(x+y+2\right)^2+2\left(y+0,5\right)^2+2012,5\ge2012,5\)

Dấu "=" xảy ra khi:

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2=0\\y+0,5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)

Vậy \(minP=2012,5\) khi \(\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)

Ta có:

\(Q=-x^2+4x-3y^2+6y+2017\)

\(=-\left(x^2-4x+4\right)-3\left(y^2-2y+1\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\le2024\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(maxQ=2024\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a: \(0.5xy\left(8y-8x\right)-6y\left(y-x\right)-4xy^2+6xy\)

\(=4xy^2-4x^2y-6y^2+6xy-4xy^2+6xy\)

\(=-4x^2y+12xy-6y^2\)

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

Thay \(\left(1\right)\)vào \(S\),ta được :

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)

\(\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2+10z+25\right)=0\)

\(\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z+5\right)^2=0\)

\(\left[{}\begin{matrix}2x-y-z=0\\y-3=0\\z+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\y=3\\z=-5\end{matrix}\right.\)

còn phần tính S bạn xem bạn có chép sai đề ko nha