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Số nào lớn hơn:
\(\left(\frac{2006-2005}{2006+2005}\right)^2hay\frac{2006^2-2005^2}{2006^2+2005^2}\)
Theo tính chất của phân thức ta có:
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006-2005}{2006+2005}.\frac{2006-2005}{2006+2005}< \frac{2006^2-2005^2}{\left(2006+2005\right)^2}\)
\(=\frac{2006^2-2005^2}{2006^2+2.2006.2005+2005^2}< \frac{2006^2-2005^2}{2006^2+2005^2}\)
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{1}{\left(2006+2005\right)^2}<\frac{4011}{2006^2+2005^2}=\frac{2006^2-2005^2}{2006^2+2005^2}\)
Ta có :
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{\left(2006-2005\right)^2}{\left(2006+2005\right)^2}=\frac{2006^2-2.2006.2005+2005^2}{2006^2+2.2006.2005+2005^2}=\frac{2006^2-2005^2}{2006^2+2005^2}\)
Vậy \(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006^2-2005^2}{2006^2+2005^2}\)
TA CÓ A= \(\left(\frac{2006-2005}{2006+2005}\right)^2\)=\(\frac{1}{4011^2}\)
B=\(\frac{2006^2-2005^2}{2006^2+2005^2}\) = \(\frac{\left(2006-2005\right)\left(2006+2005\right)}{\left(2006+2005\right)^2-2.2005.2006}\) = \(\frac{4011}{4011^2-2.2006.2005}\)
VÌ 1.(\(4011^2\)-2.200.2005)<\(4011^2\).4011 (DO \(4011^2\)>\(4011^2\)-2.2006.2005)
\(\Rightarrow\)\(\frac{1}{4011^2}\)< \(\frac{4011}{4011^2-2.2005.2006}\) .HAY A<B
VẬY A<B
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
