so sánh các số sau:
a, \(\frac{2009}{2010}\) và \(\frac{2010}{2011}\)
b, \(\frac{1}{3^{400}}\) và \(\frac{1}{4^{300}}\)
e, \(\frac{2008}{2008.2009}\) và \(\frac{2009}{2009.2010}\)
GIÚP TỚ NHÉ CÁC BẠN !
1_so sánh: \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}\) và \(\sqrt{2008}+\sqrt{2009}\)
2_ Cho biểu thức \(P=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{2010}\). CMR: \(B>86\)
Các bạn làm giúp mik với.....then kiu các bạn nhìu nhé....
so sánh:
\(\frac{2009}{2010}và\frac{2010}{2011}\)\(\frac{1}{3^{400}}và\frac{1}{4^{300}}\)
\(\frac{200}{201}+\frac{201}{202}và\frac{200+201}{201+202}\)\(\frac{2008}{2008.2009}và\frac{2009}{2009.2010}\)
2009/2010=1-1/2010<1-1/2011=2010/2011
vậy 2009/2010<2010/2011
3^400=(3^4)^100=81^100>64^100=4^300
=>1/3^400<1/4^300
Vậy 1/3^400<1/4^300
So sánh :
A=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)và B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)
Do 20092010- 2 < 20092011- 2 ⇒ B < 1
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}<\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(1+2009^{2009}\right)}{2009\left(1+2009^{2010}\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=A\Rightarrow\)B < A
So sánh
A= \(\frac{9^{2008}+1}{9^{2009}+1}\)và \(\frac{9^{2009}+1}{9^{2010}+1}\)
áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)
Ta có : \(B=\frac{9^{2009}+1}{9^{2010}+1}< 1\)
\(\Rightarrow B< \frac{9^{2009}+1+8}{9^{2010}+1+8}\)
\(\Rightarrow B< \frac{9.\left(9^{2008}+1\right)}{9.\left(9^{2009}+1\right)}=\frac{9^{2008}+1}{9^{2009}+1}\)
Vậy B < A
B = 92009 + 1/92010 + 1 < 1
=> B < 92009 + 1 + 8 / 92010 + 1 + 8 = 92009 + 9 / 92010 + 9 = 9 (92008 + 1 ) / 9 ( 92007 + 1) = A
=>B < A
#Hoq chắc _ Baccanngon
\(\frac{9^{2008}+1}{9^{2009}+1}=\frac{9^{2009}+9}{9^{2010}+9}>\frac{9^{2009}+1}{9^{2010}+1}\)
\(B=\frac{\frac{2008}{2011}+\frac{2009}{2010}+\frac{2010}{2009}+\frac{2011}{2008}+\frac{2012}{503}}{\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}}\)
so sánh 1/3400 và 1/4300
2009/2010 và 2010/2011
So sánh A và B bằng cách so sánh với 1:
\(A=\frac{2010}{2011}+\frac{2011}{2012}\)và \(B=\frac{2010+2011}{2011+2012}\)
ta có :
\(B=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
=> \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)
hay A>B
1/ So sánh
a/\(\frac{2009}{2010}\)và \(\frac{2010}{2011}\) b/\(\frac{1}{3^{400}}\)và \(\frac{1}{4^{300}}\)
c/\(\frac{200}{201}\)+ \(\frac{201}{202}\)và \(\frac{200+201}{201+202}\) d/\(\frac{2008}{2008.2009}\)và \(\frac{2009}{2009.2010}\)
a, Ta có\(\)\(\frac{2009}{2010}< \frac{2009}{2011}\)
Mà \(\frac{2009}{2011}< \frac{2010}{2011}\)
Vậy\(\frac{2009}{2010}< \frac{2010}{2011}\)
Ta có :\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)
\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)
Vì\(\frac{1}{81^{100}}< \frac{1}{64^{100}}\)
Vậy\(\frac{1}{3^{400}}< \frac{1}{4^{300}}\)
c, Ta có : B=\(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+202}\)
\(\Rightarrow\frac{200}{201}>\frac{200}{201+202}\)
\(\frac{201}{202}>\frac{201}{201+202}\)
Vậy A>B
d, Ta có \(A=\frac{2008}{2008\times2009}=\frac{1}{2019}\)
\(B=\frac{2009}{2009\times2010}=\frac{1}{2010}\)
Vì \(\frac{1}{2009}>\frac{1}{2010}\)
Vậy A>B
so sánh các số:
a.\(\frac{2009}{2010}\)và \(\frac{2010}{2011}\)
b.\(\frac{1}{3^{400}}\)và \(\frac{1}{4^{300}}\)
c.\(\frac{200}{201}\)+\(\frac{201}{202}\)và \(\frac{200+201}{201+202}\)
d.\(\frac{2008}{2008.2009}\)và \(\frac{2009}{2009.2010}\)
nghịch đảo 2 phân số ta có: \(\frac{2010}{2009}v\text{à}\frac{2011}{2010}\)
phân tích ra ta có:\(\frac{2010}{2009}=1+\frac{1}{2009}\)
\(\frac{2011}{2010}=1+\frac{1}{2010}\)
Vì \(\frac{1}{2009}>\frac{1}{2010}\)
nên \(\frac{2009}{2010}
a/ Do : 2009/2010 > 2009/2011, 2009/2011 < 2010/2011 nên 2009/2010 < 2010/2011