Đừng có md,làm không chịu làm đăng lên đây Tuấn ak

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 

https://h.vn/hoi-dap/question/604792.html

Bn tham khảo tại đây nhé !

___G-Dragon___

Xem đáp án tại đây nhé:

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

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

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

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

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

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

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

\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)

\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)

\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)

\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)

Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:

\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)

\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)