\(ad=bc=>\frac{a}{c}=\frac{b}{d}=>\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)

=>\(\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)

=>\(\frac{a^{2003}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)

\(ad=bc\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)

\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)

\(\Rightarrow\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

a) \(\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{\left(kb\right)^{2004}-b^{2004}}{\left(kb\right)^{2004}+b^{2004}}=\frac{k^{2004}b^{2004}-b^{2004}}{k^{2004}b^{2004}+b^{2004}}=\frac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(1)

\(\frac{c^{2004}-d^{2004}}{d^{2004}+d^{2004}}=\frac{\left(kd\right)^{2004}-d^{2004}}{\left(kd\right)^{2004}+d^{2004}}=\frac{k^{2004}d^{2004}-d^{2004}}{k^{2004}d^{2004}+d^{2004}}=\frac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(2)

Từ (1) và (2) => đpcm

b) \(\frac{a^{2005}}{b^{2005}}=\frac{\left(kb\right)^{2005}}{b^{2005}}=\frac{k^{2005}b^{2005}}{b^{2005}}=k^{2005}\)(1)

\(\frac{\left(a-c\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left(kb-kd\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left[k\left(b-d\right)\right]^{2005}}{\left(b-d\right)^{2005}}=\frac{k^{2005}\left(b-d\right)^{2005}}{\left(b-d\right)^{2005}}=k^{2005}\)(2)

Từ (1) và (2) => đpcm

Có : 2004A = 2004^2004+2004/2004^2004+1 = 1 + 2003/2004^2004+1

2004B = 2004^2005+2004/2004^2005+1 = 1 + 2003/2004^2005+1 < 1 + 2003/2004^2004+1 = 2014A

=> A > B

Tk mk nha

\(B=\frac{2004^{2004}+1}{2004^{2005}+1}< \frac{2004^{2004}+1+2003}{2004^{2005}+1+2003}=\frac{2004^{2004}+2004}{2004^{2005}+2004}=\frac{2004\left(2004^{2003}+1\right)}{2004\left(2004^{2004}+1\right)}=\frac{2004^{2003}+1}{2004^{2004}+1}=A\)

Vậy A > B

tớ có cách khác cũng ra kết quả giống bạn

a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(4k^4+5\right)}=\dfrac{b^4}{d^4}\)(1)

\(\dfrac{a^2b^2}{c^2d^2}=\dfrac{k^2b^2b^2}{k^2d^2d^2}=\dfrac{b^4}{d^4}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)

b.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (1)

\(\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{4b^4k^4+5b^4}{4d^4k^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(k^4+5\right)}=\dfrac{b^4}{d^4}\\\dfrac{a^2b^2}{c^2d^2}=\dfrac{bk^2b^2}{dk^2d^2}=\dfrac{k^2b^4}{k^2d^4}=\dfrac{b^4}{d^4}\end{matrix}\right.\)

Vậy.....

\(\left\{{}\begin{matrix}\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\\\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\end{matrix}\right.\)

Vậy....

Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^4}{c^4}=\dfrac{b^4}{d^4}=\dfrac{4a^4}{4c^4}=\dfrac{5b^4}{5d^4}=\dfrac{4a^4+5b^4}{4c^4+5d^4}\left(1\right)\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2b^2}{b^4}=\dfrac{c^2d^2}{d^4}=\dfrac{a^2b^2}{c^2d^2}=\dfrac{b^4}{d^4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)(đpcm)
b/ Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\left(1\right)\)
\(\Rightarrow\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)

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