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Câu hỏi của lê thị hương giang - Toán lớp 8 | Học trực tuyến

\(2x+2y+z=4\Rightarrow z=4-2x-2y\)

Ta có: \(A=2xy+yz+xz\)

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

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

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

  \(\Rightarrow2A=8y-4xy-4y^2+8x-4x^2\)

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

               \(=-4x^2-2.x.2\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)

               \(=-\left[4x^2+2.x.2\left(y-2\right)+\left(y-2\right)^2\right]+\left(y-2\right)^2-4y^2+8y\)

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

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

                     \(=-\left(2x+y-2\right)^2-3\left(y^2-2.\frac{2}{3}y+\frac{4}{9}-\frac{4}{9}-\frac{4}{3}\right)\)       

                      \(=-\left(2x+y-2\right)^2-3\left(y-\frac{2}{3}\right)^2+\frac{16}{3}\)

                        \(=\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\)

Vì \(\left(2x+y-2\right)^2\ge0;\left(y-\frac{2}{3}\right)^2\ge0\) Nên \(\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\le\frac{16}{3}\)

\(\Rightarrow A\le\frac{16}{3}:2=\frac{8}{3}\)

Dấu "=" xảy ra <=>\(\hept{\begin{cases}y-\frac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-y+2}{2}\\y=\frac{2}{3}\\z=4-2x-2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{cases}}}\)

Vậy A_Max = 8/3 khi và chỉ khi x = y = 2/3 và z = 4/3

2x + 2y + z = 4(1)
A = 2xy + yz + xz(2)
(1) z=2c<=>x+y=2-c($)
(2)<=>2xy+2yc+2cx=A
A=2B<=>xy +(x+y).c=B
xy=B-c(2-c)
($:%)=> ton tai nghiem x,y
(c-2)^2≥4[B+c(c-2)]
c^2-4c+4≥4B+4c^2-8c
-3c^2+4c≥4B-4
-3(c^2-2.2/3c+4/9)≥4B-4-4/3
-3(c-2/3)^2≥4B-16/3
=> B≤4/3
A≤8/3
dang thuc khi c=2/3; z=1/3
x=y=2/3

A=2xy+yz+xzA=2xy+yz+xz

=2xy+y(4−2x−2y)+x(4−2x−2y)=2xy+y(4−2x−2y)+x(4−2x−2y)

=−2x2−2xy+4x−2y2+4y=−2x2−2xy+4x−2y2+4y

=[−(x2+2xy+y2)+83(x+y)−169]−(x2−43x+49)−(y−43y+49)+83=[−(x2+2xy+y2)+83(x+y)−169]−(x2−43x+49)−(y−43y+49)+83=−(x+y−43)2−(x−23)2−(y−23)2+83≤83=−(x+y−43)2−(x−23)2−(y−23)2+83≤83

Vậy Amax=83Amax=83 tại 

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Bn tham khảo tại đây nhé !

___G-Dragon___

\(z=4-2x-2y\)

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

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

\(A=-2\left(x^2+\frac{y^2}{4}+1+xy-2x-y\right)-\frac{3}{2}\left(y^2-\frac{4}{3}y+\frac{4}{9}\right)+\frac{8}{3}\)

\(A=-2\left(x+\frac{y}{2}-1\right)^2-\frac{3}{2}\left(y-\frac{2}{3}\right)^2+\frac{8}{3}\le\frac{8}{3}\)

\(\Rightarrow A_{max}=\frac{8}{3}\) khi \(\left\{{}\begin{matrix}x=\frac{2}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{matrix}\right.\)