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Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2\left[\left(ax^3+3ax^2\right)+\left(3ax+9a\right)\right]\)
\(=2a\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)
\(=2a\left(x+3\right)\left(x^2+3\right)\)
2ax3 + 6ax2 + 6ax + 18a
= 2a( x3 + 3x2 + 3x + 9 )
= 2a[ ( x3 + 3x2 ) + ( 3x + 9 ) ]
= 2a[ x2( x + 3 ) + 3( x + 3 ) ]
= 2a( x + 3 )( x2 + 3 )
\(ax^2-3axy+bx-3by\\ =x\left(ax+b\right)-3y\left(ax+b\right)\\ =\left(x-3y\right)\left(ax+b\right)\)
\(5x^2y+5xy^2-a^2x-a^2y\\ =5xy\left(x+y\right)-a^2\left(x+y\right)\\ =\left(5xy-a^2\right)\left(x+y\right)\)
\(2ax^3+6ax^2+6ax+18a\\ =2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\)
\(10xy^2-5by^2+2ax-ab\\ =5y^2\left(2x-b\right)+a\left(2x-b\right)\\ =\left(5y^2+a\right)\left(2x-b\right)\)
\(ax-bx+cx-3a+3b-3c\\ =x\left(a-b+c\right)-3\left(a-b+c\right)\\ =\left(x-3\right)\left(a-b+c\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
2x^5-6x^4-2a^2x^3-6ax^3
=(2x^5-2a^2x^3)-(6x^4+6ax^3)
=2x^3(x^2-a^2)-6x^3(x+a)
=2x^3(x-a)(x+a)-6x^3(x+a)
=(x+a)(2x^4-2x^3a-6x^3)
=(x+a) 2x^3 (x-a-3)
\(=x^2\left(a+b\right)-6x\left(a+b\right)+9\left(a+b\right)\)
\(=\left(a+b\right)\left(x^2-6x+9\right)\)
\(=\left(a+b\right)\left(x-3\right)^2\)
bn làm sai rồi kìa bn
-6x nhân với b là ra âm 6bx rùi bn
mà đề cho là dương 6bx
A = 10ax - 5ay - 2x + y
= ( 10ax - 5ay ) - ( 2x - y )
= 5a( 2x - y ) - ( 2x - y )
= ( 2x - y )( 5a - 1 )
B = 2x2 - 6xy + 5x - 15y
= 2x( x - 3y ) + 5( x - 3y )
= ( x - 3y )( 2x + 5 )
C = ax2 - 3axy + bx - 3by
= ( ax2 + bx ) - ( 3axy + 3by )
= x( ax + b ) - 3y( ax + b )
= ( ax + b )( x - 3y )
D = 2ax3 + 6ax2 + 6ax + 18a
= 2ax2( x + 3 ) + 6a( x + 3 )
= ( x + 3 )( 2ax2 + 6a )
= ( x + 3 )2a( x2 + 3 )
E = 5x2y + 5xy2 + a2x + a2y ( đã sửa 1 dấu '-' )
= 5xy( x + y ) + a2( x + y )
= ( x + y )( 5xy + a2 )
F = 10xy2 - 5by2 + 2a2x - aby ( xem lại đề chứ không phân tích được :)) )
a) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=2a\left(x+3\right)\left(x^2+3\right)\)
a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
