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\(a)8x^6-27y^3=\left(2x^2\right)^3-\left(3y\right)^3=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
\(b)\left(x+3\right)^3-8=\left(x+3\right)^3-2^3\)
\(=\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+4\right]\)
\(=\left(x+1\right)\left(x^2+6x+9+2x+6+4\right)\)
\(=\left(x+1\right)\left(x^2+8x+19\right)\)
\(c)x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(d)x^3+12x^2+48x+64=x^3+3x^2\cdot4+3x\cdot16+4^3\)
\(=\left(x+4\right)^3\)
3)(9a)2-(5a-3b)2
= (9a-5a+3b)(9a+5a-3b)
= (4a+3b)(14a-3b)
a) 4.(2a-b)2-16(a-b)2
= [2(2a-b)]2 - [4(a-b)]2
= [2(2a-b)-4(a-b)].[2(2a-b)+4(a-b)]
= [4a-2b-4a+4b].[4a-2b+4a-4b]
= 2b.(8a-6b)
b) 8x3-27y3
= (2x)3 - (3y)3
= (2x - 3y).[(2x)2+2x.3y+(3y)2]
= (2x-3y)(4x2+6xy+9y2)
c) 1/64x6-125y3
= (1/4x2)3 - (5y)3
= (1/4x2 - 5y)[(1/4x2)2 + 1/4x2.5y + (5y)2]
= (1/4x2 - 5y)(1/16x4 + 5/4x2y +25y2)
d) (x+3)3-8
= (x+3-2)[(x+3)2+(x+3).2+22]
= (x+1)(x2+6x+9+2x+6+4)
= (x+1)(x2+8x+19)
e) x6+1
= (x2)3 + 13
= (x2 + 1)[(x2)2 - x2 + 1]
= (x2 + 1)(x4-x2+1)
g) x9 + 1
= (x3)3 + 13
= (x3 + 1 )[(x3)2 - x6 + 1]
= (x+1)(x2+x+1)(x6-x6+1)
= (x+1)(x2+x+1)
Mình gõ hơi lâu mới làm được nhiêu đó thôi
\(h)x^3+12x^2+48x+64=\left(x+4\right)^3\)
\(i)27-27m+9m^2-m^3=\left(3-m\right)^3\)
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
b) \(64x^3+1=\left(4x+1\right)\left(16x^2-4x+1\right)\)\
c) \(x^3y^6z^9-125=\left(xy^2z^3-5\right)\left(x^2y^4z^6+5xy^2z+25\right)\)
d) \(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)
e) \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
64x3 + 1
= ( 4x )3 + 1
= ( 4x + 1 ) ( 16x2 - 4x + 1 )
Hằng đẳng thức 6 : A3 + B3
27x6 - 8x3
= ( 3x2)3 + ( 2x )3
= ( 3x + 2x ) ( 9x2 - 6x + 4x2 )
HĐT 6
x6 - y6
= ( x2 )3 - ( y2 )3
= ( x2 - y2 ) ( x4 + x2y2 + y4 )
HĐT 7 : A3 - B3
x3y6z9 + 1
= ( xy2z3)3 + 1
= ( xy2z3 + 1 ) ( x2y4z6 + zy2z3 + 1 )
HĐT 6
a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)
= xy(z+1) +(z+1)+z(x+y)+(x+y)
= (z+1)(xy+1)+(x+y)(Z+1)
=(z+1)(xy+1+x+y)
8x3 - 27y3 = 23 . x3 - 33 . y3 = ( 2x )3 - ( 3y )3 = ( 2x - 3y ) [(2x)2 + 12xy + (3y)2 ].
Phân tích đa thức thành nhân tử ( phương pháp dùng hằng đẳng thức )
3) x6 - y6
= (x3)2 - (y3)2
= (x3 - y3).(x3 + y3)