mấy bạn chuyên toán giải giùm mk bài b) giùm ạ, mk đaq rất cần

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

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

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

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

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

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

Tham khảo:

Cho 3 số thức x,y,z thỏa mãn \(x\ge1;y\ge4;z\ge9\) tìm giá trị lớn nhất của biết thức Q=\(\dfrac{yz\sqrt{x-1}+zx\sqrt... - Hoc24

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

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

Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)

<=>\(x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+yz+zx\right)\)<=>\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

<=>\(3^2\ge3\left(xy+yz+zx\right)\)<=>\(P=xy+yz+zx\le3\)=>P_max=3 <=> x=y=z=1

Ta có BĐT đúng sau:

x2 + y2 + z2 >= xy + yz + zx

<=> (x + y + z)2 >= 3(xy + yz + zx)

<=> 9 >= 3 P <=> P <=3 (dấu bằng khi x = y = z =1)

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