2xy+6x=y−22xy+6x=y−2

⇔2x(y+3)=y+3−5⇔2x(y+3)=y+3−5

⇔(2x−1)(y+3)=−5⇔(2x−1)(y+3)=−5

Xet U(-5) nhé bạn

cảm ơn bạn nhìu nha

2xy - 6x + y = - 7

2xy - 2x.3 + y = - 7

2x(y - 3) + y = - 7

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

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

=> 2x + 1 và y - 3 là ước của - 10

=> Ư(- 10) = { ± 1; ± 2; ± 5 ± 10 }

Vì 2x + 1 là số lẻ => 2x + 1 = { ± 1; ± 5 }

Nếu 2x + 1 = 5 thì y - 3 = - 2 => x = 2 thì y = 1

Nếu 2x + 1 = 1 thì y - 3 = - 10 => x = 0 thì y = - 7

Nếu 2x + 1 = - 1 thì y - 3 = 10 => x = - 1 thì y = 13

Nếu 2x + 1 = - 5 thì y - 3 = 2 => x = - 3 thì y = 5

Vậy ( x;y ) = { ( 2;1 ); ( 0;-7 ); ( -1;13 ); ( -3;5 ) }

\(a,2x^2+y^2+6x-2xy+9=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+6x+9\right)=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-3\end{matrix}\right.\Leftrightarrow x=y=-3\\ b,A=\left(x-2021\right)^2+\left(x+2022\right)^2=x^2-4042x+2021^2+x^2+4044x+2022^2\\ A=2x^2+2x+2021^2+2022^2\\ A=2\left(x^2+x+\dfrac{1}{4}\right)+2021^2+2022^2-\dfrac{1}{2}\\ A=2\left(x+\dfrac{1}{2}\right)^2+2021^2+2022^2-\dfrac{1}{2}\ge2021^2+2022^2-\dfrac{1}{2}\\ A_{max}=2021^2+2022^2-\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)\(c,P=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+16\\ P=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+16\\ P=\left(a^2+8a+11\right)^2-16+16=\left(a^2+8a+11\right)^2\left(Đpcm\right)\)

\(2xy-y-6x-3=5\)

\(2x\left(y-3\right)-\left(y-3\right)+6=5\)

\(\Rightarrow\left(2x-1\right)\left(y-3\right)=-1\)

Với 2x-1=1=>y-3=-1

=>x=1;y=2

Với 2x-1=-1;y-3=1

=>x=0;y=4