Lời giải:
a. S Đ

b. Đ S

c. S Đ

a) Đúng

b) Sai 

c) Sai

d) Đúng 

ta có:

1/10.A=10¹⁰⁰+1/10(10⁹⁹+1)

1/10.A=10¹⁰⁰+1/10¹⁰⁰+10

1/10.A=1-(9/10¹⁰⁰+10)

1/10.B=10¹⁰¹+1/10(10¹⁰⁰+1)

1/10.B=10¹⁰¹+1/10¹⁰¹+10

1/10.B=1-(9/10¹⁰¹+10)

vì(10¹⁰¹+10)>(10¹⁰⁰+1)=>  9/10¹⁰¹+10 < 9/10¹⁰⁰+10 => 1-(9/10¹⁰¹+10) > 1-(9/10¹⁰⁰+10)

hay 1/10.A>1/10.B

=>A>B

ta có:

1/10.A=10100+1/10(1099+1)

1/10.A=10100+1/10100+10

1/10.A=1-(9/10100+10)

1/10.B=10101+1/10(10100+1)

1/10.B=10101+1/10101+10

1/10.B=1-(9/10101+10)

vì(10101+10)>(10100+1)=>  9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)

hay 1/10.A<1/10.B

=>A<B

Đáp án dưới mới đúng nhé 

vừa mình làm nhầm

a) \(\dfrac{1}{10}=0,1\)

\(\dfrac{1}{100}=0,01\)

\(\dfrac{1}{1000}=0,001\)

\(\dfrac{1}{10000}=0,0001\)

b) \(\dfrac{84}{10}=8,4\)

\(\dfrac{225}{100}=2,25\)

\(\dfrac{6453}{100}=64,53\)

\(\dfrac{25789}{10000}=2,5789\)

CTVHS là gì vậy ??

\(A=\dfrac{10^{99}+1}{10^{100}+1}\)

\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}\)

\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)

Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)

\(\Rightarrow A>B\)

Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)

\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)

\(B< \dfrac{10^{100}+10}{10^{101}+10}\)

\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)

\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)

\(B< A\)

Ta có : \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)

\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)

\(B< \dfrac{10^{100}+10}{10^{101}+10}\)

\(B< \dfrac{10.\left(10^{99}+1\right)}{10.\left(10^{100}+1\right)}\)

\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)

Vậy \(B< A\)

\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ < \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{97}{98}.\dfrac{98}{99}< \dfrac{1}{99}\\ < \dfrac{1}{10}.\\\\ =>A< \dfrac{1}{10}\)

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)