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\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{2004+1}{1-2003}\)\(=\frac{2005}{-2002}\)
\(B=\frac{2005^3-1}{2005^2+2006}\)\(=\frac{2005-1}{1+2006}=\frac{2004}{2007}\)
\(\Rightarrow A>B\)
\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}\)
\(A=\frac{2005.\left(2004^2-2003\right)}{2004^2-2003}=2005\)
\(B=\frac{2005^3-1}{2005^2+2006}\)
\(B=\frac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=\frac{2004.\left(2005^2+2006\right)}{2005^2+2006}=2004\)
Tham khảo nhé~
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ x+2009=0
⇔ x=-2009
vậy x=-2009 là nghiệm của pt
a) ( x2 + x )2 + 4( x2 + x ) = 12
<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0
<=> ( x2 + x + 2)2 - 16 = 0
<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0
<=> ( x2 + x + 6 )( x2 + x - 2) = 0
Do : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x
=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1) + 2( x - 1) = 0
<=> ( x - 1)( x + 2 ) = 0
<=> x = 1 hoặc : x = - 2
KL.....
b) Kuroba kaito làm rùi nhé 
Bạn sửa lại đề bài câu 2) nhé ^^
2) \(a+b+c+d=0\Leftrightarrow a+b=-c-d\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-\left[c^3+d^3+3cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)
\(18x^2y^2\left(?\right)4x^2y\)
câu b)
\(\left(b\right)6x^3-9x^2=3x^2\left(x-3\right)\)
\(\left(c\right)4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
\(C=b^3+c^3+ab^2+ac^2-abc=\left(b+c\right)\left(b^2-bc+c^2\right)+a\left(b^2-bc+c^2\right)=\left(b^2-bc+c^2\right)\left(a+b+c\right)\)Vì a + b + c = 0 \(\Rightarrow\left(a+b+c\right)\left(b^2-bc+c^2\right)=0\Rightarrow C=0\)
Sửa đề\(2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2006\right)+1=A\)
Đặt \(2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2006\right)+1=A\)
Ta có:
\(A=2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=\left(2005-1\right)\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=2005\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)\)\(-\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=\left(2005^{2007}+2005^{2006}+2005^{2005}+...+2005^2+2005\right)\)\(-\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=2005^{2007}⋮2005^{2007}\left(dpcm\right)\)
Phần a thành nhân tử sẵn rồi bạn:)
b,\(x^6-9x^3+8=x^6-x^3-8x^3+8\)
\(=x^3\left(x^3-1\right)-8\left(x^2-1\right)\)
\(=x^3\left(x-1\right)\left(x^2+x+1\right)-8\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4-x^3-8x-1\right)\)
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)