25x25
2004x4+2004 +7x2004
Tính:
B=2003.(2004^2003+2004^2002+...+2004^2+2004+1)-2004^2004+5
so sánh A =2004^2003+1/2004^2004+1/ và B=2004^2004+1/2004^2005+1
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
hãy CM
\(\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
cho ad=bc
\(ad=bc=>\frac{a}{c}=\frac{b}{d}=>\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
=>\(\frac{a^{2003}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
=>\(\frac{a^{2003}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
\(ad=bc\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}\)
\(\Rightarrow\frac{a^{2004}}{c^{2004}}=\frac{b^{2004}}{d^{2004}}=\frac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\frac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\)
\(\Rightarrow\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)
\(\frac{2004^{2003}+1}{2004^{2004}+1}va\frac{2004^{2004}+1}{2004^{2005}+1}\)
tinh nhanh (-2004-2004-2004-2004).(-25)
cac ban lam giup voi
(a^2004+b^2004)^2005/(c^2004+d^2004)^2005=(a^2005-b^2005)^2004/(c^2005-d^2005)^2004
A = 2004×37+2004+2×2004+2004×59+2004 324×321−201×324−324×101−18×324
tính nhanh 2004 . 37 + 2004 + 2 * 2004 + 2004 * 59 + 2004 / 324 * 321 - 201 * 324 - 324 * 101 - 18 * 324
\(\frac{2004.37+2004+2.2004+2004.59+2004}{324.321-201.324-101.324-18.324}\)
\(=\frac{2004.\left(37+1+2+59+1\right)}{324.\left(321-201-101-18\right)}\)
\(=\frac{2004.100}{324.1}\)
\(=\frac{200400}{324}\)
\(=\frac{16700}{27}\)
cái con THANH NGÂN ngu thế
Cho A = 2004 + 20042 + 20043 + 20044 + 20045 + 20046 +............................+ 20048 + 200410 . Chứng minh rằng: A chia hết cho 2005
A = (2004 + 20042 ) + ( 20043 + 20044)+ (20045 + 20046) +............................+ (20048 + 200410)
A = 2004 ( 1 + 2004 ) + 20043 ( 1 +2004 ) + .... + 20048 ( 1+ 2004 )
A = 2004.2005 + 20043.2005 +....+20048.2005
A = 2005.( 2004 + 20042 + 20043 + 20044 + 20045 + 20046 +............................+ 20048 + 200410 )
Vậy A chia hết cho 2005
có sai đề ở chỗ 2004^8+2004^10 ko bn