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AH
Akai Haruma
Giáo viên
4 tháng 2 2024
Lời giải:
a.
$=\frac{3}{5}-\frac{7}{4}=\frac{12-35}{20}=\frac{-23}{20}$
b.
$=-(2+\frac{5}{8})=-\frac{21}{8}$
c.
$=-(\frac{1}{8}+\frac{5}{9})=-\frac{9+8.5}{8.9}=\frac{-49}{72}$
d.
$=\frac{6}{13}-\frac{14}{39}=\frac{18}{39}-\frac{14}{39}=\frac{4}{39}$
e.
$=\frac{-3}{4}+\frac{5}{7}=\frac{5}{7}-\frac{3}{4}$
$=\frac{20-21}{7.4}=\frac{-1}{28}$
KV
25 tháng 1 2024
Bài 5
1) x ∈ Ư(18) = {1; 2; 3; 6; 9; 18}
x ∈ B(4) = {0; 4; 8; 12; 16; 20; ...}
Vậy không tìm được x thỏa mãn đề bài
2) x ∈ Ư(20) = {1; 2; 4; 5; 10; 20}
x ∈ B(2) = {0; 2; 4; 6; 8; 10; 12; 14; 16; 18; 20; ...}
⇒ x ∈ {2; 4; 10; 20}
3) x ∈ B(12) = {0; 12; 24; 36; 48; ...; 96; 108; ...}
Mà 30 ≤ x ≤ 100
⇒ x ∈ {36; 48; ...; 96}
4) x ∈ Ư(150) = {1; 2; 3; 5; 6; 10; 15; 25; 30; 50; 75; 150}
Mà x ≤ 50
⇒ x ∈ {1; 2; 3; 5; 6; 10; 15; 25; 30; 50}
5) 70 ⋮ x và 168 ⋮ x
⇒ x ∈ ƯC(70; 168)
Ta có:
70 = 2.5.7
168 = 2³.3.7
⇒ ƯCLN(70; 168) = 2.7 = 14
⇒ x ∈ ƯC(70; 168) = Ư(14) = {1; 2; 7; 14}
Mà x > 10
⇒ x = 14
6) Ta có:
(1995 + 2005 + x) ⋮ 5
1995 ⋮ 5
2005 ⋮ 5
⇒ x ⋮ 5
⇒ x ∈ B(5) = {0; 5; 10; 15; 20; 25; 30; 35; 40; ...}
Mà 23 < x ≤ 35
⇒ x ∈ {25; 30; 35}
KV
25 tháng 1 2024
Bài 6
1) Do 17x2y chia hết cho 2 và 5 nên y = 0
⇒ Số đã cho có dạng: 17x20
Để 17x20 chia hết cho 3 thì (1 + 7 + x + 2 + 0) ⋮ 3
⇒ (10 + x) ⋮ 3
⇒ x ∈ {2; 5; 8}
Vậy x ∈ {2; 5; 8}; y = 0
2) Do 234xy chia hết cho 2 và 5 nên y = 0
⇒ Số đã cho có dạng: 234x0
Để 234x0 chia hết cho 9 thì (2 + 3 + 4 + x + 0) ⋮ 9
⇒ (9 + x) ⋮ 9
⇒ x ∈ {0; 9}
Vậy x ∈ {0; 9}; y = 0
3) Do 4x6y chia hết cho 2 và 5 nên y = 0
Mà x - y = 4
⇒ x = 4 + y
⇒ x = 4
Vậy x = 4; y = 0
4) Do 57x2y chia hết cho 5 nhưng không chia hết cho 2 nên y = 5
⇒ Số đã cho có dạng 57x25
Để 57x25 chia hết cho 9 thì (5 + 7 + x + 2 + 5) ⋮ 9
⇒ (19 + x) ⋮ 9
⇒ x = 8
Vậy x = 8; y = 5
NL
Nguyễn Lê Phước Thịnh
CTVHS
13 tháng 9 2025
110:
\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+\cdots+\frac{198}{2}+\frac{199}{1}\)
\(=\left(1+\frac{1}{199}\right)+\left(1+\frac{2}{198}\right)+\cdots+\left(1+\frac{198}{2}\right)+1\)
\(=\frac{200}{2}+\frac{200}{3}+\cdots+\frac{200}{200}=200\left(\frac12+\frac13+\cdots+\frac{1}{200}\right)\)
=200A
=>\(\frac{A}{B}=\frac{1}{200}\)
109:
\(100-\left(1+\frac12+\frac13+\cdots+\frac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\frac12\right)+\left(1-\frac13\right)+\cdots+\left(1-\frac{1}{100}\right)\)
\(=0+\frac12+\frac23+\cdots+\frac{99}{100}=\frac12+\frac23+\cdots+\frac{99}{100}\)
108:
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\cdots+\frac{1}{101\cdot400}\)
\(=\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\cdots+\frac{299}{101\cdot400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac12-\frac{1}{301}+\cdots+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+\cdots+\frac{1}{299\cdot400}\)
\(=\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\cdots+\frac{101}{299\cdot400}\right)\)
\(=\frac{1}{101}\left(1-\frac{1}{102}+\frac12-\frac{1}{103}+\cdots+\frac{1}{299}-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac12+\frac13+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)
Do đó: \(\frac{A}{B}=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
15 tháng 1 2024
\(a,-\dfrac{5}{7}+1+\dfrac{30}{-7}\le x\le-\dfrac{1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\\ \dfrac{-5+1.7-30}{7}\le x\le\dfrac{-1+1.2+5}{6}\\ -\dfrac{28}{7}\le x\le\dfrac{6}{6}\\ -4\le x\le1\\ Vậy:x\in\left\{-4;-3;-2;-1;0;1\right\}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
15 tháng 1 2024
\(b,\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{13}\le x\le-\dfrac{9}{14}+3+\dfrac{5}{-14}\\ \left(\dfrac{21}{13}-\dfrac{8}{13}\right)+\dfrac{7}{17}\le x\le\left(-\dfrac{9}{14}-\dfrac{5}{14}\right)+3\\ 1+\dfrac{7}{17}\le x\le-1+3\\ 1\dfrac{7}{17}\le x\le2\\ Vậy:x=2\)
AH
Akai Haruma
Giáo viên
15 tháng 1 2024
Lời giải:
$\frac{1}{50}> \frac{1}{100}$
$\frac{1}{51}> \frac{1}{100}$
.....
$\frac{1}{98}> \frac{1}{100}$
$\frac{1}{99}> \frac{1}{100}$
$\Rightarrow S> \underbrace{\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}}_{50}=\frac{1}{100}.50=\frac{1}{2}$
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
15 tháng 1 2024
\(\left(3+3^2+3^3+3^4+...+3^{99}+3^{100}\right)\\ =3.\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\\ =3.4+3^3.4+...+3^{99}.4\\ =4.\left(3+3^3+...+3^{99}\right)⋮4\left(ĐPCM\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 1 2024
Chiều rộng của mảnh vườn hình chữ nhật là:
\(\dfrac{21}{4}\) : \(\dfrac{7}{3}\) = \(\dfrac{9}{4}\) (m)
Chu vi của mảnh vườn hìn chữ nhật là:
(\(\dfrac{21}{4}\) + \(\dfrac{9}{4}\)) x 2 = 15 (m)
Diện tích của mảnh vườn hình chữ nhật là:
\(\dfrac{21}{4}\) x \(\dfrac{9}{4}\) = \(\dfrac{189}{16}\) (m2)
b; Số tiền thu được khi trồng hoa để bán trên mảnh đất hình chữ nhật đó là:
80 000 x \(\dfrac{189}{16}\) = 945 000 (đồng)
KL...
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 1 2024
Bài 5:
a, Chiều rộng mảnh vườn:
\(\dfrac{21}{4}:\dfrac{7}{3}=\dfrac{9}{4}\left(m\right)\)
Chu vi mảnh đất:
\(2\times\left(\dfrac{21}{4}+\dfrac{9}{4}\right)=15\left(m\right)\)
Diện tích mảnh đất:
\(\dfrac{21}{4}\times\dfrac{9}{4}=\dfrac{189}{16}\left(m^2\right)\)
b, Số tiền thu được khi bán hoa:
\(\dfrac{189}{16}\times80000=945000\left(đồng\right)\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 1 2024
\(\dfrac{15}{34}+\dfrac{1}{3}+\dfrac{19}{34}-\dfrac{4}{3}+\dfrac{3}{7}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{3}{7}=1-1+\dfrac{3}{7}=\dfrac{3}{7}\)
Bài 1:
a; \(\dfrac{1}{n}\); \(\dfrac{1}{n+1}\) (n > 0; n \(\in\) Z)
\(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\) = \(\dfrac{n+1-1}{n.\left(n+1\right)}\) = \(\dfrac{1}{n\cdot\left(n+1\right)}\)
⇒ \(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\) = \(\dfrac{1}{n\left(n+1\right)}\) (đpcm)
Bài 1b
A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{9}\)
A = \(\dfrac{7}{18}\)
Bài1b;
B = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + ... + \(\dfrac{3}{100.103}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{100}\) - \(\dfrac{1}{103}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{103}\)
B = \(\dfrac{102}{103}\)
Bài 2:
a; C = \(\dfrac{4}{1.3}\) + \(\dfrac{4}{3.5}\) + \(\dfrac{4}{5.7}\) + ... + \(\dfrac{4}{99.101}\)
C = 2. (\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{99.101}\))
C = 2.(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\))
C = 2.(\(\dfrac{1}{1}\) - \(\dfrac{1}{101}\))
C = 2. \(\dfrac{100}{101}\)
C = \(\dfrac{200}{101}\)
Bài 2; b
D = \(\dfrac{4}{11.16}\) + \(\dfrac{4}{16.21}\) + \(\dfrac{4}{21.16}\) + ... + \(\dfrac{4}{61.66}\)
D = \(\dfrac{4}{5}\).(\(\dfrac{5}{11.16}\) + \(\dfrac{5}{16.21}\) + \(\dfrac{5}{21.16}\) + ... + \(\dfrac{5}{61.66}\))
D = \(\dfrac{4}{5}\).(\(\dfrac{1}{11}\) - \(\dfrac{1}{16}\) + \(\dfrac{1}{16}\) - \(\dfrac{1}{21}\) + \(\dfrac{1}{21}\) - \(\dfrac{1}{16}\) + ... + \(\dfrac{1}{61}\) - \(\dfrac{1}{66}\))
D = \(\dfrac{4}{5}\).(\(\dfrac{1}{11}\) - \(\dfrac{1}{66}\))
D = \(\dfrac{4}{5}\). \(\dfrac{5}{66}\)
D = \(\dfrac{2}{33}\)
Bài 2c;
E = \(\dfrac{5}{3.7}\) + \(\dfrac{5}{7.11}\) + \(\dfrac{5}{11.15}\) + ... + \(\dfrac{5}{85.89}\)
E = \(\dfrac{5}{4}\).(\(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{85.89}\))
E = \(\dfrac{5}{4}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{85}\) - \(\dfrac{1}{89}\)
C = \(\dfrac{5}{4}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{89}\))
C = \(\dfrac{5}{4}\). \(\dfrac{86}{267}\)
C = \(\dfrac{215}{534}\)
Bài 3:
Tìm \(x\):
a; \(x\).(\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{99.100}\)) = 1
\(x\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) = 1
\(x\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)) = 1
\(x\).\(\dfrac{99}{100}\) = 1
\(x\) = 1: \(\dfrac{99}{100}\)
\(x\) = \(\dfrac{100}{99}\)
Vậy \(x\) = \(\dfrac{100}{99}\)
Bài 3; b
\(x\) - (\(\dfrac{2}{1.4}\) + \(\dfrac{2}{4.7}\) + \(\dfrac{2}{7.10}\) + ... + \(\dfrac{2}{97.100}\)) = \(\dfrac{11}{25}\)
\(x\) - \(\dfrac{2}{3}\)(\(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + ... + \(\dfrac{3}{97.100}\)) = \(\dfrac{11}{25}\)
\(x\) - \(\dfrac{2}{3}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{97}\) - \(\dfrac{1}{100}\)) = \(\dfrac{11}{25}\)
\(x\) - \(\dfrac{2}{3}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)) = \(\dfrac{11}{25}\)
\(x\) - \(\dfrac{2}{3}\). \(\dfrac{99}{100}\) = \(\dfrac{11}{25}\)
\(x\) - \(\dfrac{33}{50}\) = \(\dfrac{11}{25}\)
\(x\) = \(\dfrac{11}{25}\) + \(\dfrac{33}{50}\)
\(x\) = \(\dfrac{11}{10}\)
Vậy \(x=\dfrac{11}{10}\)
Bài 3c;
\(x\) - \(\dfrac{20}{11.13}\) - \(\dfrac{20}{13.15}\) - \(\dfrac{20}{15.17}\) - ... - \(\dfrac{20}{53.55}\) = \(\dfrac{3}{11}\)
\(x\) - 10.(\(\dfrac{2}{11.13}\) + \(\dfrac{2}{13.15}\) + \(\dfrac{2}{15.17}\) + ... + \(\dfrac{2}{53.55}\)) = \(\dfrac{3}{11}\)
\(x-10.\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(x\) - 10.(\(\dfrac{1}{11}\) - \(\dfrac{1}{55}\)) = \(\dfrac{3}{11}\)
\(x\) - 10. \(\dfrac{4}{55}\) = \(\dfrac{3}{11}\)
\(x\) - \(\dfrac{8}{11}\) = \(\dfrac{3}{11}\)
\(x\) = \(\dfrac{3}{11}\) + \(\dfrac{8}{11}\)
\(x\) = 1
Vậy \(x\) = 1
Bài 4a;
F = \(\dfrac{1}{7}\) + \(\dfrac{1}{91}\) + \(\dfrac{1}{247}\)+ ... + \(\dfrac{1}{1147}\)
F = \(\dfrac{1}{1.7}\) + \(\dfrac{1}{7.13}\) + \(\dfrac{1}{13.19}\) + ... + \(\dfrac{1}{31.37}\)
F = \(\dfrac{1}{6}\).(\(\dfrac{6}{1.7}\) + \(\dfrac{6}{7.13}\) + \(\dfrac{6}{13.19}\) + ... + \(\dfrac{6}{31.37}\))
F = \(\dfrac{1}{6}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{13}\) + \(\dfrac{1}{13}\) - \(\dfrac{1}{19}\) + ... + \(\dfrac{1}{31}\) - \(\dfrac{1}{37}\))
F = \(\dfrac{1}{6}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{37}\))
F = \(\dfrac{1}{6}\). \(\dfrac{36}{37}\)
F = \(\dfrac{6}{37}\)
Bài 4;b
G = \(\dfrac{3}{15}\) + \(\dfrac{3}{35}\) + \(\dfrac{3}{63}\) + ... + \(\dfrac{3}{99.101}\)
G = \(\dfrac{3}{2}\).(\(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\) + ... + \(\dfrac{2}{99.101}\))
G = \(\dfrac{3}{2}\).(\(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\) + ... + \(\dfrac{2}{99.101}\))
G = \(\dfrac{3}{2}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\))
G = \(\dfrac{3}{2}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{101}\))
G = \(\dfrac{3}{2}\).\(\dfrac{98}{303}\)
G = \(\dfrac{49}{101}\)
Bài 4
C = (1 + \(\dfrac{3}{5}\)).(1 + \(\dfrac{3}{12}\)).(1 + \(\dfrac{3}{21}\))...(1 + \(\dfrac{3}{9797}\))
C = \(\dfrac{8}{5}\).\(\dfrac{15}{12}\).\(\dfrac{24}{21}\)...\(\dfrac{9800}{97}\)
C = \(\dfrac{2.4}{1.5}\).\(\dfrac{3.5}{2.6}\).\(\dfrac{4.6}{3.7}\)...\(\dfrac{98.100}{97.101}\)
C = \(\dfrac{2.3.4...97.98}{1.2.3.4...97}\) \(\times\) \(\dfrac{4.5.6.7...100}{5.6.7...100.101}\)
C = \(\dfrac{98}{1}\) x \(\dfrac{4}{101}\)
C = \(\dfrac{392}{101}\)
Bài 5
A = \(\dfrac{3^2}{2.5}\) + \(\dfrac{3^2}{5.8}\) + \(\dfrac{3^2}{8.11}\)
A = \(3\times\) (\(\dfrac{3}{2\times5}\) + \(\dfrac{3}{5\times8}\) + \(\dfrac{3}{8\times11}\))
A = 3 \(\times\) (\(\dfrac{1}{2}-\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\))
A = 3 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{11}\))
A = 3 \(\times\) \(\dfrac{9}{22}\)
A = \(\dfrac{27}{22}\) > 1
B = \(\dfrac{4}{5.7}\) + \(\dfrac{4}{7.9}\) + ... + \(\dfrac{4}{59.61}\)
B = 2 \(\times\) (\(\dfrac{2}{5\times7}\) + \(\dfrac{2}{7\times9}\) + ... + \(\dfrac{2}{59\times61}\))
B = 2 \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + ... + \(\dfrac{1}{59}\) - \(\dfrac{1}{61}\))
B = 2 \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{61}\))
B = 2 \(\times\) \(\dfrac{56}{305}\)
B = \(\dfrac{112}{305}\) < 1
Vì A > 1 > B vậy A > B
Bài 5:
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ... + \(\dfrac{1}{2^{2019}}\)
2 \(\times\) A = 2 \(\times\) (\(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ... + \(\dfrac{1}{2^{2019}}\))
2 \(\times\) A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ... + \(\dfrac{1}{2^{2018}}\)
2 \(\times\) A - A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ... + \(\dfrac{1}{2^{2018}}\) - ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+ ... +\(\dfrac{1}{2^{2018}}\) + \(\dfrac{1}{2^{2019}}\))
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ... + \(\dfrac{1}{2^{2018}}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) - ...- \(\dfrac{1}{2^{2018}}\)- \(\dfrac{1}{2^{2019}}\)
A = (1 - \(\dfrac{1}{2^{2019}}\)) + (\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\)) + (\(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^2}\)) +...+ (\(\dfrac{1}{2^{2018}}\) - \(\dfrac{1}{2^{2018}}\))
A = 1 - \(\dfrac{1}{2^{2019}}\) + 0 + 0 ... + 0
A = 1 - \(\dfrac{1}{2^{2019}}\) < 1