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x2 - 4y2 + x + 2y
= ( x2 - 4y2 ) + ( x + 2y )
= ( x - 2y ) ( x + 2y ) + ( x + 2y )
= ( x + 2y ) ( x - 2y + 1 )
\(x^2+4xy+4y^2-4z^2-1-4z\)
\(=x^2+4xy+4y^2-\left(4z^2+4z+1\right)\)
\(=\left(x+2y\right)^2-\left(2z+1\right)^2\)
\(=\left(x+2y+2z+1\right)\left(x+2y-2z-1\right)\)
Bài 2:
Sửa đề: \(x^3-3x^2-10x=0\)
\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)
\(x^2+4y^2+9-4xy-6x+12y\)
\(=\left(x^2-4xy+4y^2\right)+\left(-6x+12y\right)+9\)
\(=\left(x-2y\right)^2-6\left(x-2y\right)+9\)
\(=\left(x-2y-3\right)^2\)
\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)
= ( x2 + 4xy +4y2 ) - ( 4z2 +4z +1 )
= ( x + y )2 - [ (2z)2 - 2z.1 +12)]
= ( x + y )2 - (2z+1)2
= ( x + y - 2z - 1 ).( x + y + 2z + 1 )
=\(x^2+2.x.2y+\left(2y\right)^2-\left[\left(2z\right)^2+2.2z.1+1^2\right]=\left(x+2y\right)^2-\left(2z+1\right)^2=\left(x+2y+2z+1\right)\left(x+2y-2z-1\right)\)
\(=x^2\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-4\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
mik nghĩ là sai đề r
X2 + 4y2 + 2( 3x + 6y + 2xy) +9
=x^2 +4y^2 +6x+ 12y+4xy+9
=x^2 +4xy+(2y)^2 +6(x+2y)+9
=(x+2y)^2+6(x+2y)+9
=(x+2y)(x+2y+6)+9
như vậy thì số 9 sẽ bị lẻ
\(13,5x5,8-8,3x4,2-5,8x8,3+4,2x13,5\)
\(=13,5x\left(5,8+4,2\right)-8,3x\left(4,2+5,8\right)\)
\(=13,5x10-8,3x10\)
\(=135-83\)
\(=52\)
\(x^2+4xy-3x+4y^2-6y\)
\(=x^2+4xy+4y^2-3x-6y\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)
\(x^2+4xy-3x+4y^2-6y\)
\(=x^2+4xy+4y^2-3x-6y\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)