\(=\left(2x^2+xy-3x\right)-\left(22xy+11y^2-33y\right)+\left(2x+y-3\right)\)

\(=x\left(2x+y-3\right)-11y\left(2x+y-3\right)+\left(2x+y-3\right)\)

\(=\left(x-11y+1\right)\left(2x+y-3\right)\)

Hình như đề bài hơi sai sai í bn '^' Mk nghĩ sửa 12x thành 12xy nha

TL :

4x² - 12xy + 5y²

= 4x² - 2xy - 10xy + 5y²

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

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

Thế này là xog òi đó bn :)

Nếu đề bài là đúng như bn vt trên kia thỳ xl ạ :>

Mk k phân tích đc :)

Đồng ý với bạn Linh nhé :3

4x² - 12xy + 5y²

= 4x² - 2xy - 10xy + 5y²

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

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

Mk nghĩ như v ms đúng :)

\(x^2-4y^2-12y-9\)

\(=x^2-\left(4y^2+12y+9\right)\)

\(=x^2-[\left(2y\right)^2+2.2x.3+3^2]\)

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

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

\(=1\left(x^4+4\right)\)

Ta có \(x^4+4=\left(x^2\right)^2+2^2=\left(x^2+2\right)^2-2.x^2.2=\left(x^2+2\right)^2-\left(2x\right)^2\)

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

ta có :

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

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

x²-5x+5y-y²

=(x²-y²)-(5x-5y)

=(x-y)(x+y)-5(x-y)

=(x-y)(x+y-5)

\(3x-4x^2+7=-\left(4x^2-3x-7\right)=-\left(x+1\right)\left(4x-7\right)\)

  3x-4x²+7

=-4x²+3x+7

=-4x²-4x+7x+7

=-(4x²+4x)+(7x+7)

=-4x(x+1)+7(x+1)

=(x+1)(-4x+7)

\(3x-4x^2+7\)

\(=4x^2-4x+7x-7\)

\(=-\left(4x^2+4x\right)+\left(7x+7\right)\)

\(=-4x\left(x+1\right)7\left(x+1\right)\)

\(=\left(4x+7\right)\left(x+1\right)\)

Vào đây xem nha 

https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-4x-2-2x-3y-9y-2-thanh-nhan-tu-faq289539.html

hoctot

\(x^2-4x+5y^2-10y+9=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(5y^2-10y+5\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y^2-2y+1\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y-1\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0;5\left(y-1\right)^2\ge0\) mà \(\left(x-2\right)^2+5\left(y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\5\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)