\(H=\left(3x-2y\right)^2-\left(4y-6x\right)^2-\left|xy-24\right|\)

\(H=\left(3x-2y\right)^2-\left(-2\right)^2.\left(3x-2y\right)^2-\left|xy-24\right|\)

\(H=\left(3x-2y\right)^2-4\left(2x-2y\right)^2-\left|xy-24\right|\)

\(H=-3.\left(3x-2y\right)^2-\left|xy-24\right|\)

Vì \(\hept{\begin{cases}\left(3x-2y\right)^2\ge0\forall x,y\\\left|xy-24\right|\ge0\forall x,y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3\left(3x-2y\right)^2\le0\\-\left|xy-24\right|\le0\end{cases}}\)

\(\Leftrightarrow H=-3\left(3x-2y\right)^2-\left|xy-24\right|\le0\forall x,y\)

\(\Leftrightarrow H\le0\forall x,y\)

Dấu " = " xảy ra khi và chỉ 

\(\hept{\begin{cases}\left(3x-2y\right)^2=0\\\left|xy-24\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}3x=2y\\xy=24\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2y}{3}\\\frac{2y}{3}.y=24\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{2y}{3}\\y^2=36\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=6\Leftrightarrow x=4\\y=-6\Leftrightarrow x=-4\end{cases}}\)

Vậy \(Max_H=0\Leftrightarrow\left(x;y\right)\in\left\{\left(4;6\right);\left(-4;-6\right)\right\}\)

Bạn tham khảo !!!

Gửi lại : ~~ Bạn k hiểu ạ ??

\(H=\left(3x-2y\right)^2-\left(4y-6x\right)^2-\left|xy-24\right|\)

\(H=\left(3x-2y\right)^2-\left(-2\right)^2.\left(3x-2y\right)^2-\left|xy-24\right|\)

\(H=\left(3x-2y\right)^2-4\left(3x-2y\right)^2-\left|xy-24\right|\)

\(H=-3.\left(3x-2y\right)^2-\left|xy-24\right|\)

\(\text{Vì }\hept{\begin{cases}\left(3x-2y\right)^2\ge0\forall x,y\\\left|xy-24\right|\ge0\forall x,y\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-3\left(3x-2y\right)^2\le0\\-\left|xy-24\right|\le0\end{cases}}\)

\(\Rightarrow H=-3\left(3x-2y\right)^2-\left|xy-24\right|\le0\forall x,y\)

\(\Rightarrow H\le0\forall x,y\)

\(\text{Dấu "=" xảy ra khi và chỉ khi }\)

\(\hept{\begin{cases}\left(3x-2y\right)^2=0\\\left|xy-24\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\xy=24\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{2y}{3}\\\frac{2y}{3}.y=24\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{2y}{3}\\y^2=36\end{cases}\Rightarrow\hept{\begin{cases}x=\pm4\\y=\pm6\end{cases}}}\)

\(\text{Vậy Hmax = 0 xảy ra khi (x;y) }\in\left\{\left(4;6\right);\left(-4;6\right);\left(4;-6\right);\left(-4;-6\right)\right\}\)

Học tốt

Hông hiểu? 

\(M=2019\left(x-2y\right)^{2018}-\left(6y-3y\right)^{2018}-\left|xy-2\right|\\ \)

\(Do\left(x-2y\right)^{2018}\ge0\Rightarrow2019\left(x-2y\right)^{2019}\)

\(\left(6y-3x\right)^{2018}\ge0\Rightarrow-\left(6y-3x\right)^{2018}\le0\)

\(\left|xy-2\right|\ge0\Rightarrow-\left|xy-2\right|\le0\)=>\(M\le0-0-0=0.\)

GIá tri lon nhat cua Mla 0 khi va chi khi

\(\hept{\begin{cases}x-2y=0\\6y-3x=0\\xy-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\6y=3x\\xy=2\end{cases}\Rightarrow\hept{\begin{cases}x=2y\\y=\frac{1}{2}x\\xy=2\end{cases}}}\)

\(\Rightarrow xy=2y.y=2y^2\Rightarrow y^2=1\Rightarrow y=\pm1\Rightarrow x=\pm2\)

vay ..........

ta có:

\(\left(3x-2y\right)^2\)>  0

\(\left(4y-6x\right)^2\)> 0

\(\left|xy-24\right|\)>    0

dấu "=" xảy ra (=)

\(\hept{\begin{cases}\left(3x-2y\right)^2=0\\\left(4y-6x\right)^2=0\\\left|xy-24\right|=0\end{cases}}\left(=\right)\hept{\begin{cases}3x-2y=0\\4y-6x=0\\xy-24=0\end{cases}}\)\(\)còn lại mk chưa tính ra

bạn ơi nếu làm thế này là sai đó,các biến ở các hạnh tử giống nhau mà

Ta thấy : \(-\left(3x-2y\right)^2\le0\forall x,y\)

\(-\left(4y-6x\right)^2\le0\forall x,y\)

\(-\left|xy-24\right|\le0\forall x,y\)

\(\Rightarrow-\left(3x-2y\right)^2+\left(4y-6x\right)^2+\left|xy-24\right|\le0\forall x,y\)

\(\Leftrightarrow-\left(3x-2y\right)^2+\left(4y-6x\right)^2+\left|xy-24\right|+2019\le2019\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(3x-2y\right)^2=0\\\left(4y-6x\right)^2=0\\\left|xy-24\right|=0\end{cases}}\)  \(\Leftrightarrow\hept{\begin{cases}3x=2y\\xy=24\end{cases}}\) 

Ta có : \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}=k\) \(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\end{cases}}\)

Khi đó : \(xy=2k\cdot3k=6k^2=24\)

\(\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)

Với \(k=-2\Rightarrow\hept{\begin{cases}x=-4\\y=-6\end{cases}}\) ( thỏa mãn )

Với \(k=2\Rightarrow\hept{\begin{cases}x=4\\y=6\end{cases}}\) ( thỏa mãn )

Vậy : GTLN của \(-\left(3x-2y\right)^2+\left(4y-6x\right)^2+\left|xy-24\right|+2019=2019\) tại \(\left(x,y\right)\in\left\{\left(4,6\right);\left(-4,-6\right)\right\}\)

Giups mik vs