Giải:

A=10²⁰⁰⁴+1/10²⁰⁰⁵+1

10A=10²⁰⁰⁵+10/10²⁰⁰⁵+1

10A=10²⁰⁰⁵+1+9/10²⁰⁰⁵+1

10A=1+9/10²⁰⁰⁵+1

Tương tự:

B=10²⁰⁰⁵+1/10²⁰⁰⁶+1

10B=1+9/10²⁰⁰⁶+1

Vì 9/10²⁰⁰⁵+1>9/10²⁰⁰⁶+1 nên 10A>10B

⇒A>B

Chúc bạn học tốt!

\(10A=10.\dfrac{10^{2004}+1}{10^{2005}+1}=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\\ 10B=10.\dfrac{10^{2005}+1}{10^{2006}+1}=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)

vì \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Đề bài đâu bn?

Ta có: \(10\cdot A=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\)

\(10B=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)

mà \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\)

nên 10A>10B

hay A>B

Co j góc lệch cùng trời cuối đất cú mèo

bài 1

a)<

b)>

N = what?????????????????????????????????????????????????

P/s: thiếu đề nên ko làm đc 

\(\hept{\begin{cases}\\\end{cases}45656.lll\hept{\begin{cases}\\\end{cases}}}\)

A = (-7/10²⁰⁰⁵-7/10²⁰⁰⁶) - 8/10²⁰⁰⁶

B = (-7/10²⁰⁰⁵-7/10²⁰⁰⁶) - 8/ 10²⁰⁰⁵

Vì 10²⁰⁰⁶  > 10²⁰⁰⁵ => 8/10²⁰⁰⁶ > 8/10²⁰⁰⁵ => - 8/10²⁰⁰⁶ < - 8/10²⁰⁰⁵ => A < B

\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)

\(=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

\(B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)

\(=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

VÌ \(10^{2006}\ge10^{2005}\)

\(\Rightarrow\frac{-8}{10^{2006}}\le\frac{-8}{10^{2005}}\)

\(\Rightarrow\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\le\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

\(\Rightarrow A\le B\)