Dễ thấy: \(2008^3+1>0\); \(2008^2-2007>0\)
Nên \(\frac{2008^3+1}{2008^2-2007}>0\Leftrightarrow A>0\)
và \(2009-2010< 0\); \(2009^3-1>0\)
\(\Rightarrow\frac{2009^3-1}{2009-2010}< 0\Leftrightarrow B< 0\)
Vậy A > B
1. tích các nghiệm cua phương trình(x-2009)(x-2008)(x-2007)...(x-1)x(x+1)...(x+2010)=0
2.cho abc=2008 . khi do gia tri cua bieu thuc
\(\frac{2008a}{\left(ab+2008a+2008\right)}+\frac{b}{bc+b+2008}+\frac{c}{ac+c+1}=0\)
3.gia tri cua bieu thuc Q=(3+1)(32+1)(34+1)...(33994+1)
Giải phương trình: \(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
giải phương trình: \(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2009}{2}+\frac{x+2010}{1}\)\(=\left(-2010\right)\)
\(\frac{x+1}{2011}+\frac{x+2}{2010}+\frac{x+3}{2009}=\frac{x+4}{2008}+\frac{x+5}{2007}+\frac{x+6}{2006}\)
giai phuong trinh
CAC BAN TRA LOI NHANH HO TUI NHE
Thu gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
\(B=\frac{\left(a+2008\right)!+\left(a+2009\right)!}{\left(a+2008\right)!-\left(a+2009!\right)}\)
Tìm x biết: \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
giải pt
\(a,0,25^3+x^2+x=0\)
\(b,\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
Giải phương trình:
\(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
Giải phương trình
\(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}.\)
