B1 a, x^3+1=0      <=> x^3 = -1

<=> x=-1

b, x^2=2x<=> x^2-2x = 0

<=> x.(x-2)=0               <=> x=0 hoặc x-2=0

<=> x=0 hoặc x=2

c, 3x^2-6x-24=0

<=> (3x^2+6x)-(12x+24) = 0

<=> (x+2) . (3x-12) = 0

<=> x+2=0 hoặc 3x-12=0

<=> x=-2 hoặc x=4

B2, a, Có 2012^2 = 2012.2012 = (2011+1).2012 = 2011.2012 + 2012

= 2011.2012+2011 + 1 = 2011.(2012+1) +1 = 2011.2013 +1 > 2011.2013

=> 2011.2013 < 2012^2

c, a+b+c = 0      <=> a+b=-c

<=> (a+b)^3 = -c^3

<=> a^3+b^3+3ab.(a+b) = -c^3

<=> a^3+b^3+c^3 + 3ab(a+b)=0

<=> a^3+b^3+c^3 = -3ab.(a+b) = -3ab.(-c) = 3abc => ĐPCM

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)

\(P=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+2008\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\)

Đặt \(x^2+10x+21=t\)

\(\Rightarrow P=\left(t-5\right)\left(t+3\right)+2008=t^2-2t+1993\)

\(\Rightarrow P\) chia \(x^2+10x+21\) dư \(1993\)