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

Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

A = \(\dfrac{n^9+1}{n^{10}+1}\) 

\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n -  \(\dfrac{n-1}{n^9+1}\)

B = \(\dfrac{n^8+1}{n^9+1}\)

\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) =  n - \(\dfrac{n-1}{n^8+1}\)

Vì n > 1 ⇒ n - 1> 0

       \(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)

⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)

⇒ A < B