Bài 1:

   D     =      5  + 5² + 5³+...+ 5¹⁰⁰

5.D     =             5² + 5³+...+5 ¹⁰⁰ + 5¹⁰¹

5D - D = 5¹⁰¹ - 5

4D       = 5¹⁰¹ - 5

  D      = \(\dfrac{5^{101}-5}{4}\)

Bài 2:

So sánh 

a, 54⁴ = (2.3³)⁴ = 2⁴.3¹²  

    21¹² = (3.7)¹² = 3¹².7¹²

Vì 2⁴ < 7¹² nên 54⁴ < 21¹²

b, 3³⁹ và 11²¹

    3³⁹   =   (3¹³)³

   11²¹ = (11⁷)³

     3¹³ = (3²)⁶.3 = 9⁶.3 < 9⁷ < 11⁷ 

Vậy 3³⁹  < 11²¹

1) \(D=5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=1+5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=\dfrac{5^{100+1}-1}{5-1}\)

\(\Rightarrow D+1=\dfrac{5^{101}-1}{4}\)

\(\Rightarrow D=\dfrac{5^{101}-1}{4}-1=\dfrac{5^{101}-5}{4}=\dfrac{5\left(5^{100}-1\right)}{4}\)

2)

a) \(21^{12}=\left(21^3\right)^4=9261^4>54^4\Rightarrow54^4< 21^{12}\)

b) \(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 11^{20}< 11^{21}\)

\(\Rightarrow3^{39}< 11^{21}\)

c) \(201^{60}=\left(201^4\right)^{15}=\text{1632240801}^{15}\)

\(398^{45}=\left(398^3\right)^{15}=\text{63044792}^{15}< \text{1632240801}^{15}\)

\(201^{60}>398^{45}\)

\(A=2^{21}-2^{18}=2^{18}\left(2^3-1\right)=7\cdot2^{18}=14\cdot2^{17}⋮14\\ B=3^{100}-2^{100}+3^{98}-2^{98}\\ B=3^{98}\left(3^2+1\right)-2^{97}\left(2^3+2\right)\\ B=3^{98}\cdot10-2^{97}\cdot10=10\left(3^{98}-2^{97}\right)⋮10\\ C=1+3+3^2+...+3^{99}\\ C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\\ C=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\\ C=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{96}\right)\\ C=40\left(1+3^4+...+3^{36}\right)⋮40\)

a ) 87  .   ( − 2 0 )     <   | 87  .   ( − 2 0 ) |

b ) 87  .   ( − 2 0 )   =  87  .   ( − 2 0 )

c ) 87  .   (   − 2 0 )   =     −   | −  87  .  2 0 |

Lời giải:
a.

\(\overline{a,87}+\overline{2,b2}=a+0,87+2,02+\overline{0,b}\\ =(a+\overline{0,b})+(0,87+2,02)\\ =\overline{a,b}+2,89\)

b.

\(\overline{3a,81}+\overline{4,b5}+\overline{13,9c}=30,81+a+4,05+\overline{0,b}+13,9+\overline{0,0c}\\ =(30,81+4,05+13,9)+(a+\overline{0,b}+\overline{0,0c})\\ =48,76+\overline{a,bc}=\overline{a,bc}+20,36+28,4> \overline{a,bc}+20,36+28,04 \)

87.(-20) < |87. (-20)|

87.(-20) = - |-87. 20|

87.(-20) = 87.(-20)

a. Quy đồng: \(BCNN\left(92,107\right)=9844\)

 \(\Rightarrow\dfrac{87}{92}=\dfrac{9309}{9844};\dfrac{102}{107}=\dfrac{9384}{9844}\)

Do \(\dfrac{9309}{9844}< \dfrac{9384}{9844}\)

Vậy: \(\dfrac{87}{92}< \dfrac{102}{107}\)

b. Quy đồng: \(BCNN\left(60,17\right)=1020\)

\(\Rightarrow\dfrac{-61}{60}=\dfrac{-1037}{1020};\dfrac{16}{17}=\dfrac{960}{1020}\)

Do: \(\dfrac{-1037}{1020}< \dfrac{960}{1020}\)

Vậy: ...