Phân tích đa thứ sau thành nhân tư
a) x^6+x^4+x^2.y^2+y^4-y^6
b) 9(a+b)^2+16a-4b-(a-b)^2-24
c) (2a+b)^3+6a+3b-4
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1) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
Phân tích đa thức thành nhân tử
a. 3ab ( x+ y) - 6ab ( y+ x)
=( x + y) ( 3ab - 6ab )
= ( x +y ) ( - 3ab)
b.7a (x - 3)+a2(x2 - 9)
=7a( x- 3) + a2 ( x2 - 32)
=7a ( x - 3 ) + a2 ( x- 3 ) ( x+3 )
= ( x- 3) . 7a + a2 ( x + 3)
= ( x- 3) ( 7a +a2x + 3a2)
c. 34 (x + y) -x -y
= 34 ( x+ y) - ( x+y)
=(x +y ) ( 34 - 1) = 33 ( x+ y)
d. 25 x4 - 942
=( 5x2 )2 - 942
=( 5x2 - 94 ) ( 5x2+94)
e.( 5a - b )2 - ( 2a +3b)2
=( 5a -b -2a - 3b) (5a -b + 2a + 3b)
=(3a - 4b) (7a+ 2b)
k. 22 -3a - b2 +3b
=( 22 - b2 ) + ( -3a +3b)
=( 2-b) (2+b) + 3( -a +b)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
a) \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=x^6-y^6+x^4+x^2y^2+y^4\)
\(=\left[\left(x^3\right)^2-\left(y^3\right)^2\right]+x^4+2x^2y^2+y^4-x^2y^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)^2-\left(x-y\right)\left(x+y\right)x^2y^2+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x^2+y^2\right)^2\left[\left(x-y\right)\left(x+y\right)+1\right]-x^2y^2\left[\left(x-y\right)\left(x+y\right)+1\right]\)
\(=\left[\left(x-y\right)\left(x+y\right)+1\right]\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2-y^2+1\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\)
c) \(\left(2a+b\right)^3+6a+3b-4\)
\(=\left(2a+b\right)^3+3\left(2a+b\right)-4\)
Đặt 2a + b = t.
Ta có: \(t^3+3t-4\)
\(=t^3-t^2+t^2-t+4t-4\)
\(=t^2\left(t-1\right)+t\left(t-1\right)+4\left(t-1\right)\)
\(=\left(t-1\right)\left(t^2+t+4\right)\)
Thay t = 2a + b vào biểu thức:
\(\left(t-1\right)\left(t^2+t+4\right)=\left(2a+b-1\right)\left(4a^2+4ab+b^2+2a+b+4\right)\)