Ta có :
\(\left(20^{2006}+11^{2006}\right)^{2007}=20^{2006.2007}+2.20^{2006}.11^{2006}+11^{2006.2007}\)
\(\left(20^{2007}+11^{2007}\right)^{2006}=20^{2007.2006}+2.20^{2007}.11^{2007}+11^{2007.2006}\)
Vì \(2.20^{2006}.11^{2006}< 2.20^{2007}.11^{2007}\) nên \(\left(20^{2006}+11^{2006}\right)^{2007}< \left(20^{2007}+11^{2007}\right)^{2006}\)
Chúc bạn học tốt ~
So sánh (20^2006+11^2006)^2007 và (20^2007+11^2007)^2006
Hãy so sánh:
\((20^{2006}+11^{2006})^{2007}\)và \((20^{2007}+11^{2007})^{2006}\)
Mọi người giúp mình với mình đang cần gấp!!!
so sánh :
a.3^300 +4^300 và 3.24^100
b.(20^2006 + 11^2006)^2007 và (20^2007 +11^2007)^2006
c.(1/2^2-1).(1/3^2-1).(1/4^2-1)..........(1/1000^2-1) và -1/2
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)Chứng minh:
a)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}^{ }}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
b)\(\left(4a+5b\right)\left(7c-11d\right)=\left(7a-11b\right)\left(4c+5d\right)\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
Hãy so sánh:
a) \(3^{300}\)+ \(4^{300}\)và \(3.24^{100}\)
b) \((20^{2006}+11^{2006})^{2007}\)và \((20^{2007}+11^{2007})^{2006}\)
c) \(3333^{4444}\)và \(4444^{3333}\)
So sánh hai số sau:
\(A=\frac{1}{2006}\)
\(B=\frac{1}{2008}+\left(\frac{1}{2008}+\frac{1}{2008^2}\right)^2+...+\left(\frac{1}{2008}+\frac{1}{2008^2}+...+\frac{1}{2008^{2007}}\right)^{2007}\)
So sánh hai số sau:
\(A=\frac{1}{2006}\)
\(B=\frac{1}{2008}+\left(\frac{1}{2008}+\frac{1}{2008^2}\right)^2+...+\left(\frac{1}{2008}+\frac{1}{2008^2}+...+\frac{1}{2008^{2007}}\right)^{2007}\)
Chứng minh rằng
B=\(\left[1.2.3.4.5.6...2006\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\right]⋮2007\)
