\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(=2\cdot\dfrac{3}{16}\)
\(=\dfrac{3}{8}\)
\(A:x=759\)
\(\dfrac{3}{8}:x=759\)
\(\Rightarrow x=\dfrac{3}{8}:759=\dfrac{1}{2024}\)
#AvoidMe

A=2(1/20+1/30+...+1/240)

=2(1/4-1/5+1/5-1/6+...+1/15-1/16)

=2*3/16=3/8

A:x=759

=>x=3/8:759=1/2024

1/2A=1/20+1/30*1/42+...+1/240
1/2A=1/4*5+1/5*6+1/6*7+...+1/15*16
1/2A=1/4-1/5+1/5-1/6+1/6-1/7+...1/15-1/16
1/2A=1/4-1/16=3/16
A=3/16:1/2 A=3/8
Ta có: A: x =759
        3/8: x =759
                x=3/8 : 759 = 1/2024

a: m⊥AB

n⊥AB

Do đó: m//n

A =   1 − 1 2 . 1 − 1 3 . 1 − 1 4 ... 1 − 1 19 . 1 − 1 20 A = 1 2 . 2 3 . 3 4 .... 18 19 . 19 20 A = 1 20 > 1 21

Vậy A> 1 21

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)

Ta có :

\(\frac{1}{10}< 1\)

\(\frac{1}{15}< 1\)

\(\frac{1}{21}< 1\)

........................

\(\frac{1}{120}< 1\)

\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)

\(\Rightarrow A< 1\)( đpcm)

Ta có : A =  \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)

= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)

= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)

= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)

= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)

= \(2\times\frac{3}{16}\)

= \(\frac{3}{8}\)< 1 

=> A < 1 

A < 1 nhe ban

Lời giải:

$A=1+3+3^2+3^3+....+3^{2026}$

$=1+3+3^2+(3^3+3^4+3^5+3^6)+(3^7+3^8+3^9+3^{10})+....+(3^{2023}+3^{2024}+3^{2025}+3^{2026})$

$=13+3^2(3+3^2+3^3+3^4)+3^6(3+3^2+3^3+3^4)+...+3^{2022}(3+3^2+3^3+3^4)$

$=13+(3^2+3^6+...+3^{2022})(3+3^2+3^3+3^4)$

$=13+(3^2+3^6+...+3^{2022}).120$

$\Rightarrow A$ chia $120$ dư $13$

A=(1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁶)+...+(3¹¹⁷+3¹¹⁸+3¹¹⁹+3¹²⁰)

A=(1+3+3²+3³)+3⁴.(1+3+3²+3³)+...+3¹¹⁷.(1+3+3²+3³)

A=40+3⁴.40+3⁷.40+...+3¹¹⁷.40

A=40.(1+3⁴+3⁷+...+3¹¹⁷) : 40 =1+3⁴+3⁷+..+3¹¹⁷ và dư 0

cảm ơn

\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)

\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)

\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)

\(A=3+3^2+3^3+...+3^{2012}\)

\(A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+...+3^{2012}\right)\)

\(A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(A=3.40+3^5.40+...+3^{2009}.40\)

\(A=120+3^4.120+...+3^{2008}.120\)

\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)

2) 
a = 135k + 88 = 120k + 15k + 88 (k là số tự nhiên) 
a chia 120 dư 58 ---> 15k + 88 chia 120 dư 58 ---> 15k + 30 chia hết cho 120 
Vì a là số bé nhất nên chọn k = 6 ---> a = 135.6 + 88 = 898. 

a = 135k + 88 = 120k + 15k + 88 (k là số tự nhiên) 
a chia 120 dư 58 ---> 15k + 88 chia 120 dư 58 ---> 15k + 30 chia hết cho 120 
Vì a là số bé nhất nên chọn k = 6 ---> a = 135.6 + 88 = 898. 

a = 135k + 88 = 120k + 15k + 88 (k là số tự nhiên) 
a chia 120 dư 58 ---> 15k + 88 chia 120 dư 58 ---> 15k + 30 chia hết cho 120 
Vì a là số bé nhất nên chọn k = 6 ---> a = 135.6 + 88 = 898.