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Bài 1 : Phân tích các đa thức sau thành nhân tử :
a) 8x3 - 64
=(2x)3 + 43
=(2x+4)(4x2 - 8x + 16)
c) 125x3 + 1
=5x3 + 13
=(5x+1)(25x2 +5x+1)
d) 8x3 - 27
=(2x)3 - 33
=(2x - 3)(2x2 + 6x + 9)
e) 1 + 8x6y3
=1 + (2x2y)3
=(1 + 2x2y)(4x4y2 -2x2y + 1)
f) 125x3 + 27y3
=(5x)3 + (3y3)
=(5x + 3y)(25x2 - 15xy + 9y2)
Bài 1
a) \(8x^3-64\)
\(=\left(2x\right)^3-4^3\)
\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)
c) \(125x^3+1\)
\(=\left(5x\right)^3+1^3\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)
\(=\left(2x\right)^3-3^3\)
\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)
e) \(1+8x^6x^3\)
\(=1^3+\left(2x^2y\right)^3\)
\(=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)
f) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
a) (x+2) \(\left(x^2-2x+4\right)\)
b) (3 - 2y) \(\left(9+6y+4y^2\right)\)
d) (4x - y) \(\left(16x^2+4xy+y^2\right)\)
bạn giải chi tiết hộ mình với nha.
mk sắp phải nộp bài r. huhuhuhu
1. x3 + 8 = (x + 2 )(x2 - x + 1)
2. 27 - 8y3 = ( 3 - 2y ) ( 9 + 6y + 4y2 )
3. y6 + 1 = (y2)3 + 1 = ( y2 + 1) ( y4 - y2 +1 )
4.64x3 - \(\dfrac{1}{8}\)y3 = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x2 + 2xy + \(\dfrac{1}{4}\)y2)
5. 125x6 - 27y9 = (5x2)3 - (3y3)3
= ( 5x2 - 3y3)(25x4 +15x2y3 + 9y6)
a) x3 - 1 + 5x2 - 5 + 3x - 3
= x3 + 5x2 + 3x - 9
= x3 + 6x2 - x2 + 9x - 6x - 9
= ( x3 + 6x2 + 9x ) - ( x2 + 6x + 9 )
= x( x2 + 6x + 9 ) - ( x2 + 6x + 9 )
= ( x2 + 6x + 9 )( x - 1 )
= ( x + 3 )2( x - 1 )
b) a5 + a4 + a3 + a2 + a + 1
= ( a5 + a4 + a3 ) + ( a2 + a + 1 )
= a3( a2 + a + 1 ) + 1( a2 + a + 1 )
= ( a2 + a + 1 )( a3 + 1 )
= ( a2 + a + 1 )( a + 1 )( a2 - a + 1 )
c) x3 - 3x2 + 3x - 1 - y3
= ( x3 - 3x2 + 3x - 1 ) - y3
= ( x - 1 )3 - y3
= ( x - 1 - y )[ ( x - 1 )2 + ( x - 1 )y + y2 ]
= ( x - 1 - y )( x2 - 2x + 1 + xy - y + y2 )
d) 5x3 - 3x2y - 45xy2 + 27y3
= ( 5x3 - 45xy2 ) - ( 3x2y - 27y3 )
= 5x( x2 - 9y2 ) - 3y( x2 - 9y2 )
= ( 5x - 3y )( x2 - 9y2 )
= ( 5x - 3y )[ x2 - ( 3y )2 ]
= ( 5x - 3y )( x - 3y )( x + 3y )
Ta có
a, a3+b3=(a+b)(a2-ab+b2)
b,x6+1=(x2+1)(x2-x+1)
c,125x3-1/8=(5x)3-1/23=(5x-1/2)(25x2+2,5x+1/4)
d,0,001x3-27y3=(0,1)3-(3y)3=(0,1-3y)(0,01+0,3y+y2)
k cho mính nha