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\(x^4-6x^3+54x-81\)
\(=x^4+3x^3-9x^3+27x^2-27x^2 +81x-27x-81\)
\(=\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27+81\right)\)
\(=x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)
\(=\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)
\(=\left(x+3\right)\left(x-3\right)^3\)
a) 16x2-(x2+4)2= (4x)2-(x2+4)2
= (4x-x2-4)(4x+x2+4)
\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)
= (3x-2)3
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)
=2y(x2+2xy+y2+x2-y2+x2-2xy+y2)
= 2y(3x2+y2)
\(x^4+x^3+2x^2+x+1=x^4+x^2+x^3+x+x^2+1\)
\(=x^2\left(x^2+1\right)+x\left(x^2+1\right)+1\left(x^2+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
cái cuối là \(\left(x^2+1\right)\left(x^2+x+1\right)\)
Lời giải:
$x^3-4x^2-12x+27$
$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$
$=x^2(x+3)-7x(x+3)+9(x+3)$
$=(x+3)(x^2-7x+9)$
x4+x3+2x2+x+1=x4+x3+x2+x2+x+1=(x4+x3+x2)+(x2+x+1)
=x2(x2+x+1)+(x2+x+1)
=(x2+x+1)(x2+1)
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
(x^+1)*(x^2+1+x0
A, m2-7m+12=m2-3m+12-4m=m(m-3)-4(m-3)=(m-4)(m-3)
B, 2x4-x3+27-54x=x3(2-x)-27(2-x)=(x3-27)(2-x)=(x-3)(x2+3x+9)(2-x)
a) m^2 -7m +12 = m^2 -3m -4m +12
=m(m -3)-4 (m- 3)
=(m-4)(m-3)
b) 2x^4-x^3 -54x+ 27
=(2m^4-x^3)- (54x - 27)
=x^3(2x-1)-27(2x-1)
=(x^3-27)(2x-1)