\(C=-x^2+2xy-4y^2+2x+10y-3\)

\(=-\left(x^2-2xy+y^2\right)+3y^2+2x+10y-3\)

\(=-\left(x^2-2xy+y^2\right)+2x-2y+12y-3y^2-3\)

\(=-\left(x^2-2xy+y^2\right)+2\left(x-y\right)-3y^2+12y-3\)

\(=-\left(x^2-2xy+y^2\right)+2\left(x-y\right)-\left(3y^2-12y+12\right)+9\)

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

Xét \(-\left(x^2-2xy+y^2\right)+2\left(x-y\right)-3.\left(y^2-4y+4\right)+9\le9\)

\(\Rightarrow Max_C=9\)

\(D=-5\left(x^2+\dfrac{4}{5}x+\dfrac{1}{5}\right)\)

\(=-5\left(x^2+2\cdot x\cdot\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{1}{25}\right)\)

\(=-5\left(x+\dfrac{2}{5}\right)^2-\dfrac{1}{5}< =-\dfrac{1}{5}\)

Dấu = xảy ra khi x=-2/5

Ta có : \(x^2+2xy-4y^2-2x+10y-8\)

\(=x^2+2.x.\left(y-1\right)+\left(y-1\right)^2-4y^2+10y-8-\left(y-1\right)^2\)

\(=\left(x+y-1\right)^2-5y^2+12y-9\)

....

Đặt A=\(-x^2+2x\left(y+1\right)-\left(y-1\right)^2-3y^2+8y+6\)

=\(-\left(x-y+1\right)^2-3\left(y^2-\frac{8}{3}y+\frac{16}{9}\right)+\frac{34}{3}\)

=\(-\left(x-y+1\right)^2-3\left(y-\frac{4}{3}\right)^2+\frac{34}{3}\le\frac{34}{3}\)

dấu = xảy ra khi \(\left\{\begin{matrix}x-y+1=0\\y-\frac{4}{3}=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=\frac{1}{3}\\y=\frac{4}{3}\end{matrix}\right.\)

Vậy max A=\(\frac{34}{3}\)khi và chỉ khi x=1/3, y=4/3

1) a) Đặt biểu thức là A

\(A=2x^2+4y^2-4xy-4x-4y+2017\)

\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)

\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)

\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)

Vậy: Min_A=2008 khi x=-3; y=-2

3) a) \(A=\dfrac{1}{x^2+x+1}\)

\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)

Vậy Min_A là \(\dfrac{4}{3}\) khi x=-0,5

a/ Đề sai, hệ số của \(y^2\) phải âm thì biểu thức mới tồn tại max

b/ \(B=-3x^2-9x-7=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)

\(B_{max}=-\frac{1}{4}\) khi \(x=-\frac{3}{2}\)

c/ \(C=-\left(x^2+y^2+1-2xy-2x+2y\right)-3\left(y^2-4y+4\right)+5\)

\(C=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\)

\(C_{max}=5\) khi \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

a: \(M=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)

\(=-2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}\right)\)

\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\forall x\)

Dấu '=' xảy ra khi x=3/4

b: Tham khảo:

*Trả lời:

a) Có vẻ như đề sai nên mình sửa lại:

\(2x^2y+2xy^2-x-y=\left(2x^2y+2xy^2\right)-\left(x+y\right)=2xy\cdot\left(x+y\right)-\left(x+y\right)=\left(2xy-1\right)\left(x+y\right)\)

b) \(8x^3-12x^2+6x-1=\left(2x\right)^3-3\cdot4x^2+3.2x-1=\left(2x-1\right)^3\)

c)\(4x^2-4xy+y^2-9=\left(4x^2-4xy+y^2\right)-9=\left(2x-y\right)^2-3^2=\left(2x-y-3\right)\left(2x-y+3\right)\)

e)\(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2y+y^2=\left(5x^2-y\right)^2\)

h)\(x^2-7xy+10y^2=x^2-2xy-5xy+10y^2=\left(x^2-2xy\right)-\left(5xy-10y^2\right)=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-5y\right)\left(x-2y\right)\)