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

c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)

\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)

100^100+1<100^101+1

=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)

=>100C>100D

=>C>D

b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)

\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)

2020^2022+1>2020^2021+1(Do 2022>2021)

=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)

=>2020E<2020F

=>E<F

a) Ta có:

\(\begin{array}{l}\frac{{123}}{7} = \frac{{123.4}}{{7.4}} = \frac{{492}}{{28}}\\17,75 = \frac{{1775}}{{100}} = \frac{{71}}{4} = \frac{{71.7}}{{4.7}} = \frac{{497}}{{28}}\end{array}\)

Vì 492 < 497 nên \(\frac{{492}}{{28}} < \frac{{497}}{{28}}\) hay \(\frac{{123}}{7} < 17,75\)

b) Ta có:

\(\begin{array}{l} - \frac{{65}}{9} = \frac{{( - 65).8}}{{9.8}} = \frac{{ - 520}}{{72}}\\ - 7,125 = \frac{{ - 7125}}{{1000}} = \frac{{ - 57}}{8} = \frac{{ - 57.9}}{{8.9}} = \frac{{ - 513}}{{72}}\end{array}\)

Vì 520 > 513 nên -520 < -513. Do đó, \(\frac{{ - 520}}{{72}} < \frac{{ - 513}}{{72}}\) hay \( - \frac{{65}}{9}\) < -7,125

a) Tổng \({S_n}\) là tổng của cấp số nhân có số hạng đầu \({u_1} = 1\) và công bội \(q = \frac{1}{3}\) nên ta có:

\({S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{1 - q}} = \frac{{1\left( {1 - {{\left( {\frac{1}{3}} \right)}^n}} \right)}}{{1 - \frac{1}{3}}} = \frac{{1 - {{\left( {\frac{1}{3}} \right)}^n}}}{{\frac{2}{3}}} = \frac{3}{2}\left( {1 - \frac{1}{{{3^n}}}} \right) = \frac{3}{2} - \frac{1}{{{{2.3}^{n - 1}}}}\)

b) Ta có:

\(\begin{array}{l}{S_n} = 9 + 99 + 999 + ... + \underbrace {99...9}_{n\,\,chu\,\,so\,\,9} = \left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + ... + \left( {\underbrace {100...0}_{n\,\,chu\,\,so\,\,0} - 1} \right)\\ = \left( {10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,so\,\,0}} \right) - n\end{array}\)

Tổng \(10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,so\,\,0}\) là tổng của cấp số nhân có số hạng đầu \({u_1} = 10\) và công bội \(q = 10\) nên ta có:

\(10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,s\^o \,\,0} = \frac{{10\left( {1 - {{10}^n}} \right)}}{{1 - 10}} = \frac{{10 - {{10}^{n + 1}}}}{{ - 9}} = \frac{{{{10}^{n + 1}} - 10}}{9}\)

Vậy \({S_n} = \frac{{{{10}^{n + 1}} - 10}}{9} - n = \frac{{{{10}^{n + 1}} - 10 - 9n}}{9}\)