Anh/ chị viết rõ đề bằng công thức toán được không ạ?

Vd : 1/2(2x+2y+z)^2 là \(\frac{1}{2\left(2x+2y+z\right)^2}\) hay sao?

\(P=8x^3+8y^3+\frac{z^3}{\left(2x+2y+2z\right)\left(4xy+2yz+2zx\right)}\) đúng ko ạ?

\(\frac{2}{8x-4x^2-5}\)

Xét mẫu:    \(8x-4x^2-5=-4x^2+8x-4-1=-\left(4x^2-8x+4\right)-1=-\left(2x-2\right)^2-1\)

Vì  \(-\left(2x-2\right)^2\le0\Rightarrow-\left(2x-2\right)^2-1\le-1\)

 Nên  \(\frac{2}{8x-4x^2-5}\le\frac{2}{-1}\le-2\)

Vậy giá trị lớn nhất của \(\frac{2}{8x-4x^2-5}\)là-2

toán 12 nha

A = 9x² - 6xy + 5y² + 1 = (3x)² + 2.3y + y² + (2y)² + 1 = ( 3x + y)² + ( 2y )² +1 
mà ( 3x + y)² > 0 và ( 2y )² > 0 

=> ( 3x + y )² + (2y)² + 1 > 0

Vậy gtnn của A là 1 

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)