Phân tích các đa thức sau thành nhân tử:
a/ (x2 + x)2 + 9x2 + 18x + 20
b/ x4 - 2x2 - 400x - 9999
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\(\left(x^2+6x-1\right)^2+2x^2+x^4+2\left(x^2+6x-1\right)\left(x^2+1\right)\)
\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+\left(x^2+1\right)^2-1=\left(x^2+6x-1+x^2+1\right)^2-1=\left(2x^2+6x\right)^2-1=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)
\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+x^4+2x^2\)
\(=\left(x^2+6x-1\right)\left(x^2+6x-1+2x^2+2\right)+x^4+2x^2\)
\(=\left(x^2+6x-1\right)\left(3x^2+6x+1\right)+x^4+2x^2\)
\(=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)
\(=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)\left(x^2+x+1\right)\)
\(x^4+x^3+2x^2+x+1\)
\(=x^4+x^3+x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^2+1\right)\)
\(=\left(x^6+2x^5+x^4\right)-2\left(x^5+2x^4+x^3\right)+2\left(x^4+2x^3+x^2\right)\)
\(=x^2\left(x^2+x\right)^2-2x\left(x^2+x\right)^2+2\left(x^2+x\right)^2\)
\(=\left(x^2+x\right)^2\left(x^2-2x+2\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)
\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\\ =4\left(x^2+16x+60\right)\left(x^2+17x+60\right)-3x^2\)
Đặt \(x^2+16x+60=a\)
\(=4a\left(a+x\right)-3x^2\\ =4a^2+4ax-3x^2\\ =\left(2a-x\right)\left(2a+3x\right)\\ =\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\\ =\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)
(x2+15x+50)(x2+18x+72)−3x2=4(x+5)(x+10)(x+6)(x+12)−3x2=4(x2+16x+60)(x2+17x+60)−3x24(�2+15�+50)(�2+18�+72)−3�2=4(�+5)(�+10)(�+6)(�+12)−3�2=4(�2+16�+60)(�2+17�+60)−3�2
Đặt x2+16x+60=a�2+16�+60=�
=4a(a+x)−3x2=4a2+4ax−3x2=(2a−x)(2a+3x)=[2(x2+16x+60)−x][2(x2+16x+60)+3x]=(2x2+31x+120)(2x2+35x+120)
a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
b) \(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2y+2\right)\left(3x+2y\right)\)
a: \(5x^2-4\left(x^2-2x+1\right)-5\)
\(=5x^2-4x^2+8x-4-5\)
\(=x^2+8x-9\)
\(=\left(x+9\right)\left(x-1\right)\)
b: \(9x^2+6x-4y^2+4y\)
\(=\left(3x+2y\right)\left(3x-2y\right)+2\left(3x+2y\right)\)
\(=\left(3x+2y\right)\left(3x-2y+2\right)\)
a, \(x^2-4x+3=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
TH1 : x = 3 ; TH2 : x = 1
b, \(2x^2-3x-2=0\Leftrightarrow\left(x-2\right)\left(x+\frac{1}{2}\right)=0\)
TH1 : x = 2 ; TH2 : x = -1/2
c, Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2+2t-8=0\Leftrightarrow\left(t-2\right)\left(t+4\right)=0\)
TH1 : t = 2 ; TH2 : t = -4
Tương tự ...
1a)
x2 - 4x + 3 = x2 - x - 3x + 3
= x( x - 1 ) - 3( x - 1 )
= ( x - 1 )( x - 3 )
2c)
2x2 - 3x - 2 = 2x2 + x - 4x - 2
= x( 2x +1 ) - 2( 2x + 1 )
= ( 2x + 1 )( x - 2 )
3e)
x4 + 2x2 - 8 (*)
Đặt t = x2
(*) <=> t2 + 2t - 8
= t2 - 2t + 4t - 8
= t( t - 2 ) + 4( t - 2 )
= ( t - 2 )( t + 4 )
= ( x2 - 2 )( x2 + 4 )
4b) x2 + 4x - 12 = x2 - 2x + 6x - 12
= x( x - 2 ) + 6( x - 2 )
= ( x - 2 )( x + 6 )
d) 2x3 + x - 2x2 - 1 = 2x2( x - 1 ) + 1( x - 1 )
= ( x - 1 )( 2x2 + 1 )
f) x2 - 2xy - 3y2 = ( x2 - 2xy + y2 ) - 4y2
= ( x - y )2 - ( 2y )2
= ( x - y - 2y )( x - y + 2y )
= ( x - 3y )( x + y )
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
\(2x^2+x-6\)
\(=2x^2-3x+4x-6\)
\(=x\left(2x-3\right)+2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(x+2\right)\)