\(A=\frac{2004x37+2004+2x2004+2004x59+2004}{324x321-201x324-324x101-18x324}\)
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NN
1
TC
9
4 tháng 7 2018
\(A=\frac{2004x37+2004+2x2004+2004x59+2004}{324x321-201x324-324x101-18x324}\)
\(A=\frac{2004x\left(37+1+2+59+1\right)}{324x\left(321-201-101-18\right)}\)
\(A=\frac{2004x\left[\left(37+1+2\right)+\left(59+1\right)\right]}{324x\left[321-\left(201+101+18\right)\right]}\)
\(A=\frac{2004x\left(40+60\right)}{324x\left[321-320\right]}\)
\(A=\frac{2004x100}{324x1}=\frac{2004100}{324}=\frac{16700}{27}\)
VN
1
XO
29 tháng 9 2019
Ta có : \(\frac{2004\times37+2004+2\times2004+2004\times49+2004}{324\times321-201\times324-324\times101-18\times324}\)
\(=\frac{2004\times\left(37+1+2+1+49\right)}{324\times\left(321-201-101-18\right)}\)
\(=\frac{2004\times90}{324\times1}=\frac{2004\times90}{324}=\frac{1670}{3}\)
26 tháng 10 2020
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
LC
25 tháng 10 2015
\(ad=bc=>\frac{a}{c}=\frac{b}{d}=>\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
=>\(\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
=>\(\frac{a^{2003}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
25 tháng 10 2015
\(ad=bc\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)
\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
\(\Rightarrow\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)
22 tháng 7 2018
\(B=\frac{2003+2004}{2004+2005}=\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)
Ta có: \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)
\(\frac{2004}{2005}>\frac{2004}{2004+2005}\)
\(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2005}\)
\(A>B\)
Vậy A>B
\(A=\frac{106700}{27}\)