\(x^2-5x+2xy-10y\)

\(=x^2+2xy+y^2-y^2-5x-10y\)

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

\(=x\left(x+2y\right)-5\left(x+2y\right)\)

\(=\left(x-5\right)\left(x+2y\right)\)

xong!

cảm ơn nha

k cho mk đi

\(=2xy+5x+4y^2+10y\)

\(=x\left(2y+5\right)+2y\left(2y+5\right)\)

\(=\left(x+2y\right)\left(2y+5\right)\)

\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)

\(A=-\left(x^2+y^2+9-2xy+6x-6y\right)-4\left(x^2+2x+1\right)-\left(y^2-4y+4\right)+16\)

\(A=-\left(x-y+3\right)^2-4\left(x+1\right)^2-\left(y-2\right)^2+16\le16\)

\(A_{max}=16\) khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

=(x^5-5x^2)-(2xy+10y)

=(x^5-5x^2)-(2xy-10y)

=x^2.(x-5)-2y.(x-5)

=(x^2-2y).(x-5)

k nha

\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)

\(=x^2y\left(x-5\right)-2y\left(x-5\right)+0\)

\(=\left(x-5\right)\left(x^2y-2y\right)\)

xong phân tích nốt cái bậc 2 kia để được max điểm :)))

\(x^2-5x+2xy-10y\)

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

\(=x\left(x-5\right)+2y\left(x-5\right)\)

\(=\left(x+2y\right)\left(x-5\right)\)

=2x³y²-10x²y+20xy²