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a) x4+2x2+1=(x2+1)2
b)=3x2(a+b)+x(a+b)+5(a+b)=(a+b)(3x2+x+5)
c)=x2(a-b)-2x(a-b)-3(a-b)=(a-b)(x2-2x-3)=(a-b)(x-3)(x+1)
d)=2x(y2-a2)-5by(y+a)=(y+a)(2xy-2xa-5by)
\(\text{a) x}^4+2x^2+1=\left(x^2+1\right)^2\)
\(\text{b) 3}ax^2+3bx^2+ãx+bx+5a+5b=\left(3ax^2+3bx^2\right) +\left(ax+bx\right)+\left(5a+5b\right)=3x^2+x\left(a+b\right)+5\left(a+b\right)=\left(a+b\right)\left(3x^2+x+5\right)\)
\(\text{c) a}x^2-bx^2-2ax+2bx-3a+3b=\left(\text{a}x^2-bx^2\right)-\left(2ax-2bx\right)-\left(3a-3b\right)=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)=\left(x^2-2x-3\right)\left(a-b\right)\)
b: \(=2x^2-2x-5x+5\)
\(=\left(x-1\right)\left(2x-5\right)\)
\(a,=x\left(x^2-4\right)+ax\left(x-2\right)\\ =x\left(x-2\right)\left(x+2\right)+ax\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+ax\right)\\ =x\left(x+a+2\right)\left(x-2\right)\\ b,=2x^2-2x-5x+5\\ =2x\left(x-1\right)-5\left(x-1\right)\\ =\left(2x-5\right)\left(x-1\right)\\ c,=\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)\\ =\left(x+3\right)\left(x^2-2x+6\right)\)
Lời giải:
$x^2-y^2+a^2-b^2+2ax+2by=(x^2+a^2+2ax)-(y^2+b^2-2by)$
$=(x+a)^2-(y-b)^2=(x+a-y+b)(x+a+y-b)$
\(ax^2+a-axy+2ax-ay\)
\(a\left(x^2+2x+1\right)-ay\left(x+1\right)\)
\(a\left(x+1\right)^2-ay\left(x+1\right)\)
\(\left(x+1\right)\left[a\left(x+1\right)-ay\right]\)
\(\left(x+1\right)\left(ax+a-ay\right)\)
\(a\left(x+1\right)\left(x+1-y\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
= (8x^2-2ax) + (4xy-ay)
= 2x.(4x-a) + y.(4x-a) = (2x+y).(4x-a)
k mk nha
1: \(-x^2+2x+8\)
\(=-\left(x^2-2x-8\right)\)
\(=-\left(x-4\right)\left(x+2\right)\)
2: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
2 x 2 + 3 x + 5 = 2 x 2 - 2 x + 5 x - 5 = 2 x 2 - 2 x + 5 x - 5 = 2 x x - 1 + 5 x - 1 = x - 1 2 x + 5
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 )
\(18-2x^2-2ax-6a\)
\(=2.\left(3^2-x^2\right)-2a\left(x+3\right)\)
\(=2.\left(3-x\right).\left(3+x\right)-2a.\left(x+3\right)\)
\(=\left(x+3\right).2.\left(3-x-a\right)\)
Tham khảo nhé~