\(E=-8x^2-3y^2-26x+6y+100\)
\(E=\left(-8x^2-26x\right)+\left(-3y^2+6y\right)+100\)
\(E=\left(-8x^2-13x-13x-169+169\right)+\left(-3y^2+3y+3y-9+9\right)+100\)
chết chết sr mình nhấn lộn đang định nhấn nút huỷ
E=-8(x^2+13x/4+169/64)-3(y^2-2y+1)+607/8
=-8(x+13/8)^2-3(y-1)^2+607/8
=> GTLN là 607/8
\(E=-8x^2-3y^2-26x+6y+100\)
\(E=\left(-8x^2-26x\right)+\left(-3y^2+6y\right)+100\)
\(E=-8.\left(x^2+3,25x\right)-3.\left(y^2-2y\right)+100\)
\(E=-8.\left(x^2+1,625x+1,625x+\dfrac{169}{64}-\dfrac{169}{64}\right)-3.\left(y^2-y-y-1+1\right)+100\)
\(E=-8.\left[\left(x^2+1,625x\right)+\left(1,625x+\dfrac{169}{64}\right)-\dfrac{169}{64}\right]-3.\left[\left(y^2-y\right).\left(y-1\right)+1\right]+100\)
\(E=-8.\left[x.\left(x+1,625\right)+1,625.\left(x+1,625\right)-\dfrac{169}{64}\right]-3.\left[y.\left(y-1\right).\left(y-1\right)+1\right]+100\)
\(E=-8.\left[\left(x+1,625\right)^2-\dfrac{169}{64}\right]-3.\left[\left(y-1\right)^2+1\right]+100\)
\(E=-8.\left(x+1,625\right)^2+\dfrac{169}{8}-3.\left(y-1\right)^2-3.\left(y-1\right)^2-3+100\)
\(E=-8.\left(x+1,625\right)^2-3.\left(y-1\right)^2+\dfrac{945}{8}\)
\(E=-\left[8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\right]+\dfrac{945}{8}\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(8.\left(x+1,625\right)^2\ge0;3.\left(y-1\right)^2\ge0\)
\(\Rightarrow8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\ge0\)
\(\Rightarrow-\left(8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\right)\le0\)
\(-\left(8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\right)+\dfrac{945}{8}\le\dfrac{945}{8}\)
Hay \(E\le\dfrac{945}{8}\) với mọi giá trị của \(x;y\in R\)
Để \(E=\dfrac{945}{8}\) thì \(-\left(8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\right)+\dfrac{945}{8}=\dfrac{945}{8}\)
\(\Rightarrow-\left(8.\left(x+1,625\right)^2+3.\left(y-1\right)^2\right)=0\)
\(\Rightarrow8.\left(x+1,625\right)^2+3.\left(y-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}8.\left(x+1,625\right)^2=0\\3.\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x+1,625\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+1,625=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1,625\\y=1\end{matrix}\right.\)
Vậy giá trị nhỏ nhất của biểu thức E là \(\dfrac{945}{8}\) đạt được khi và chỉ khi \(x=-1,625;y=1\)
Chúc bạn học tốt nha!!!
E max <=> -E min <=>\(8x^2+3y^2+26x-6y-100Min\)
\(=2\left(2x+3,25\right)^2+3\left(y-1\right)^2-124,125\)
Vì \(\left(2x+3,25\right)^2+3\left(y-1\right)^2\ge0\left(moix;y\right)\)
Nên -E\(\ge-124,125\) Do đó E \(\le124,125\)
Dấu = khi \(\left\{{}\begin{matrix}\left(2x+3,25\right)^2=0\\3\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,625\\y=1\end{matrix}\right.\)
Do đó max E = 124,125 <=> x=-1,625 và y=1
