A= 2009 mũ 10 + 2009 mũ 9= 1.071024406(thêm 33 số 0 đằng sau số này nữa) + 5.331131937(thêm 29 số 0 đằng sau số này)= 1.071557519(thêm 33 số 0)

B= 2009 mũ 10= 1.071024406(thêm 33 số 0 đằng sau số này nữa)

Nếu mik sai thì nói nhé! Mik tính bằng máy tính Casio fx-570VN PLUS đó! Nhớ k mink nha!

So sánh hai phân số : 

\(\frac{2009.2009+2008}{2009.2009+2009}\)và \(\frac{2009.2009+2009}{2009.2009+2010}\)

Dễ thấy:

\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)

\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)

=>\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

Hay \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)

Vậy A > B

ta có :

\(\frac{2008}{2009}< 1\)

\(\frac{2009}{2010}< 1\)

\(\frac{2010}{2011}< 1\)

\(\frac{2011}{2012}< 1\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 1+1+1+1\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 4\)

vậy ....................

2008/2009+2009/2010+2010/2011+2011/2008         4

=2008/2008=1                                                            4

Vì 1<4 nên 2008/2009+2009/2010+2011/2008 <   4

\(2009A=\frac{2009^{2010}+2009}{2009^{2010}+1}=\)\(\frac{2009^{2010}+1+2008}{2009^{2010}+1}=1+\frac{2008}{2009^{2010}+1}\)

\(2009B=\frac{2009^{2009}+2009}{2009^{2009}+1}=\frac{2009^{2009}+1+2008}{2009^{2009}+1}\)\(=1+\frac{2008}{2009^{2009}+1}\)

Vì \(1+\frac{2008}{2009^{2010}+1}< 1+\frac{2008}{2009^{2009}+1}\) \(\Leftrightarrow A< B\)

\(A=\frac{2009^{2009}+1}{2009^{2010}+1}\Rightarrow2009A=\frac{2009^{2010}+2009}{2009^{2010}+1}\)

\(2009A=\frac{2009^{2010}+1}{2009^{2010}+1}+\frac{2008}{2009^{2010}+1}\)

\(2009A=1+\frac{2008}{2009^{2010}+1}\)

..... sory bn mk hơi luwoif chút nên bn tự lm tương tự vs phần B và so sánh nhé!^^

chuan

Lần sau viết cái đề rõ rõ ra nhs!!!

a) \(A=2+2^2+2^3+................+2^{100}\)

\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)

\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)

\(\Rightarrow A=2^{101}-2\)

b) \(B=1+3+3^2+..................+3^{2009}\)

\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)

\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)

\(\Rightarrow2B=3^{2010}-1\)

\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)

c) \(C=4+4^2+4^3+................+4^n\)

\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)

\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)

\(\Rightarrow3C=4^{n+1}-4\)

\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)

\(A=2+2^2+2^3+...+2^{100}\)

\(2A=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2A-A=2^{101}-2\)

\(\Rightarrow A=2^{101}-2\)

Vậy \(A=2^{101}-2\).

\(B=1+3^2+3^3+...+3^{2009}\)

\(3B=3+3^3+3^4+...+3^{2010}\)

\(\Rightarrow3B-B=3^{2010}-7\)

\(\Rightarrow2B=3^{2010}-7\)

\(\Rightarrow B=\dfrac{3^{2010}-7}{2}\)

Vậy \(B=\dfrac{3^{2010}-7}{2}\).

\(C=4+4^2+4^3+...+4^n\)

\(4C=4^2+4^3+4^4+...+4^{n+1}\)

\(\Rightarrow4C-C=4^{n+1}-4\)

\(\Rightarrow3C=4^{n+1}-4\)

\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)

Vậy \(C=\dfrac{4^{n+1}-4}{3}\).