\(=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(4x-13y\right)\left(x-y\right)\)

\(4x^2-17xy+13y^2=\left(4x^2-4xy\right)+\left(13y^2-13xy\right)=4x\left(x-y\right)+13y\left(x-y\right)=\left(4x+13y\right)\left(x-y\right)\)

\(4x^2-17xy+13y^2\\ =4x^2-4xy-13xy+13y^2\\ =4x\left(x-y\right)-13y\left(x-y\right) \\ =\left(x-y\right)\left(4x-13y\right)\)

a,\(x^2-y^2+2x+1=\left(x^2-2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)b

\(4x^2-17xy+13y^2=\left(4x^2-4xy\right)+\left(13y^2-13xy\right)=4x\left(x-y\right)-13y\left(x-y\right)\)\(=\left(4x-13y\right)\left(x-y\right)\)

Nếu thấy đúng thì tick nha bạn,cảm ơn.

a) 4x² - 17xy + 13y²

=4x²-4xy-13xy+13y²

=4x(x-y)-13y(x-y)

=(x-y)(4x-13y)

b) x⁸ + x⁴ +1

=x⁸+2x⁴+1-x⁴

=(x⁴+1)²-x⁴

=(x⁴+1-x²)(x⁴+1+x²)

\(x^8+x^4+1\)

\(=x^8+2x^4+1-x^4\)

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

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

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

\(a,=\left(3-x+1\right)\left(9+3x-3+x^2-2x+1\right)\\ =\left(4-x\right)\left(x^2+x+7\right)\\ b,=4x^2-4xy-13xy+13y^2\\ =4x\left(x-y\right)-13y\left(x-y\right)\\ =\left(4x-13y\right)\left(x-y\right)\\ c,=4\left(x^2-xy-2y^2\right)\\ =4\left(x^2+xy-2xy-2y^2\right)\\ =4\left(x+y\right)\left(x-2y\right)\\ d,=x^3+4x^2+5x^2+20x+6x+24\\ =\left(x+4\right)\left(x^2+5x+6\right)\\ =\left(x+4\right)\left(x^2+2x+3x+6\right)\\ =\left(x+4\right)\left(x+2\right)\left(x+3\right)\\ f,=x\left(x+4y\right)-3\left(x+4y\right)=\left(x-3\right)\left(x+4y\right)\\ g,=4x^3+4x^2-29x^2-29x-24x-24\\ =\left(x+1\right)\left(4x^2-29x-24\right)\\ =\left(x+1\right)\left(4x^2-32x+3x-24\right)\\ =\left(x+1\right)\left(x-8\right)\left(4x+3\right)\)

\(a,27-\left(x-1\right)^3=\left(3-x+1\right)\left[9+3\left(x-1\right)+\left(x+1\right)^2\right]=\left(4-x\right)\left(9+3x-3+x^2+2x+1\right)=\left(4-x\right)\left(x^2+5x+7\right)\)

\(b,4x^2-17xy+13y^2=\left(4x^2-4xy\right)-\left(13xy-13y^2\right)=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

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

\(d,x^3+9x^2+26x+24=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(f,4xy+x^2-3x-12y=x\left(4y+x\right)-3\left(x+4y\right)=\left(x+4y\right)\left(x-3\right)\)

\(g,4x^3-25x^2-53x-24=\left(4x^3-32x^2\right)+\left(7x^2-56x\right)+\left(3x-24\right)=\left(4x^2+7x+3\right)\left(x-8\right)=\left[\left(4x^2+4x\right)+\left(3x+3\right)\right]=\left(4x+3\right)\left(x+1\right)\left(x-8\right)\)

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

\(2x^2-3ax-9a^2\)

\(=2x^2-6ax+3ax-9a^2\)

\(=2x\left(x-3a\right)+3\left(x-3a\right)\)

\(=\left(x-3a\right)\left(2x+a\right)\)

\(2x^2-17xy-9y^2\)

\(=2x^2+xy-18xy-9y^2\)

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)