\(C=1-6y-5y^2-12xy-9x^2\)

\(\Rightarrow C=-4y^2-12xy-9x^2-y^2-6y+1\)

\(\Rightarrow C=-\left(4y^2+12xy+9x^2\right)-\left(y^2+6y+9\right)+1+9\)

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

mà \(\left\{{}\begin{matrix}-\left(2y-3x\right)^2\le0,\forall x;y\\-\left(y+3\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\le10\)

\(\Rightarrow GTLN\left(C\right)=10\left(tạix=-2;y=-3\right)\)

Đặt \(A=-x^2-y^2+xy+2x+2y\)

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

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

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

anh lm chi tiết hộ e ạ 
 

-2x(x + 3) + x(2x - 1) = 10

-2x² - 6x + 2x² - x = 10

-7x = 10

x = -10/7

a) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\) (sửa \(\dfrac{x}{2}\rightarrow x^2\))

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

\(=x^6-\dfrac{1}{27}\)

b) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

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

\(=\dfrac{1}{27}x^3+8y^3\)

Lưu ý : Áp dụng hằng đẳng thức đáng nhớ \(a^3\pm b^3=...\)

Thank you