Phân tích đa thức thành nt
(x^4-625)^2-100x^2+1
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mjk sửa lại
a)100x^2 -( x^2+25)^2
=[10x-(x2+25)][10x+(x2+25)]
=(10x-x2-25)(10x+x2+25)
=-(x2-10x+25)(x+5)2
=-(x-5)2(x+5)2
b)(x+4)^3 - 64
=(x+4)3-43
=(x+4-4)[(x+4)2+(x+4).4+16]
=x(x2+8x+16+4x+16+16)
=x(x2+12x+48)
c) x^6 + y^6
=(x2)3+(y2)3
=(x2+y2)(x4+x2y2+y4)
a) 9a4+6x2+1-4a2
=(3a2+1)-4a2
=(3a2+1-2a)(3a2+1+2a) (hằng đẳng thức 3)
a/\(=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
b/ \(=x^4-7x^2+9=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
c/ \(=4x^2-2x-6x+3=2x\left(2x-1\right)-3\left(2x-1\right)=\left(2x-1\right)\left(2x-3\right)\)
d/ \(=y^4+2xy^3+2x^2y^2-2xy^3-4x^2y^2-2x^3y+2x^2y^2+4x^3y+4x^4=y^2\left(y^2+2xy+2x^2\right)-2xy\left(y^2+2xy+2x^2\right)+2x^2\left(y^2+2xy+2x^2\right)=\left(y^2+2xy+2x^2\right)\left(y^2-2xy+2x^2\right)\)
\(A=4x^4+625\)
\(=4x^4+100x^2+625-100x^2\)
\(=\left(2x^2+25\right)^2-100x^2\)
\(=\left(2x^2+5x+25\right)\left(2x^2-5x+25\right)\)
x2( x + 1 )2 + 2x2 + 2x - 8
= [ x( x + 1 ) ]2 + 2( x2 + x ) - 8
= ( x2 + x )2 + 2( x2 + x ) - 8 (*)
Đặt a = x2 + x
(*) = a2 + 2a - 8
= a2 - 2a + 4a - 8
= a( a - 2 ) + 4( a - 2 )
= ( a - 2 )( a + 4 )
= ( x2 + x - 2 )( x2 + x + 4 )
= ( x2 - x + 2x - 2 )( x2 + x + 4 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 4 )
= ( x - 1 )( x + 2 )( x2 + x + 4 )
ta co
\(x^2\left(x+1\right)^2+2x^2+2x-8\)
=\(\left(x\left(x+1\right)\right)^2+2x\left(x+1\right)+1-9\)
=\(\left(x^2+x+1\right)^2-9\)
=\(\left(x^2+x-8\right)\left(x^2+x+10\right)\)
=\(\left(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(\left(x+\frac{1}{2}\right)^2-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(x+\frac{1}{2}-\sqrt{\frac{33}{4}}\right)\left(x+\frac{1}{2}+\sqrt{\frac{33}{4}}\right)\left(x^2+x+10\right)\)