cho a+b+c=0. chứng minh: 2(a5+b5+c5)=5abc(a2+b2+c2)
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=a2-a-(a2+5a+6)
=a2-a-a2-5a-6
=-6a-6=6(-a-1) luôn chia hết cho 6
vậy...
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\(x^2-4x+13=x^2-4x+4+9\)
\(=\left(x-2\right)^2+9\)
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a/ => x3 = 64 => x3 = 43 => x = 4
b/ => 4x2 - 12x + 9 - x2 - 10x - 25 = 0
=> 3x2 - 22x - 16 = 0
=> (x - 8)(3x + 2) = 0
=> x - 8 = 0 => x = 8
hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3
Vậy x = 8 ; x = -2/3
c/ => x3 - x2 - 4x2 + 8x - 4 = 0
=> x3 - 5x2 + 8x - 4 = 0
=> (x - 2)2 (x - 1) = 0
=> (x - 2)2 = 0 => x - 2 = 0 => x = 2
hoặc x - 1 = 0 => x = 1
Vậy x = 2 ; x = 1
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a,3x3y3-15x2y2=3x2y2(xy-5)
b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)
c,(3x-1)2-16=(3x-1)2-42=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)
d,x3-3x2+3x-1=x3-1-(3x2+3x)=x3-1-3x(x+1)=(x3-1-3x)(x+1)
e,125x3+1=(5x)3+13=(5x+1)(25x2-5x.1+12)
f,x3+6x2y+12xy2+8y3=x3+3.x2.2y+3.x.(2y)2+(2y)3=(x+2y)3
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(a+b)3+(a-b)3=(a3+3a2b+3ab2+b3)+(a3-3a2b+3ab2-b3)
=a6+6a2b4