\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\frac{99}{100}\)

\(A=-\frac{98}{100}=-\frac{49}{50}\)

\(A=\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(A=\frac{1.5.6.\left(1^3+2^3+4^3+9^3\right)}{1.3.5.\left(1^3+2^3+4^3+9^3\right)}=2\)

5^x + 5^x+2 = 650

=> 5^x + 5^x.5² = 650

=> 5^x.(1 + 5²) = 650

=> 5^x.(1 + 25) = 650

=> 5^x.26 = 650

=> 5^x = 650 : 26

=> 5^x = 25 = 5²

=> x = 2

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\left(đpcm\right)\)

\(\frac{-1}{3}+\frac{0,2-0,375+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{2}.\left(\frac{2}{10}-\frac{3}{8}+\frac{5}{11}\right)}\)

\(=\frac{-1}{3}+\frac{1}{-\frac{3}{2}}\)

\(=\frac{-1}{3}+\frac{-2}{3}=-\frac{3}{3}=-1\)

B = 2² + 2³ + 2⁴ + ... + 2⁹⁸ + 2⁹⁹

2B = 2³ + 2⁴ + 2⁵ + ... + 2⁹⁹ + 2¹⁰⁰

2B - B = (2³ + 2⁴ + 2⁵ + ... + 2⁹⁹ + 2¹⁰⁰) - (2² + 2³ + 2⁴ + ... + 2⁹⁸ + 2⁹⁹)

B = 2¹⁰⁰ - 2²

B = 2¹⁰⁰ - 4

Ta có:

abc : 11 = a + b + c

=> abc = 11.(a + b + c)

=> 100a + 10b + c = 11a + 11b + 11c

=> 89a = b + 10c

=> b + 10c chia hết cho 89

Mà b;c là chữ số; b + 10c khác 0 do 89a khác 0

=> \(0< b+10c< 110\)

=> b + 10c = 89

=> b = 9; c = 8; a = 1

Vậy abc = 198

+ Nếu a < b thì a + b < b + b

=> a + b < 2.b < a.b (vì a > 2)

+ Nếu a = b thì a + b = b + b

=> a + b = 2.b < a.b (vì a > 2)

+ Nếu b > a thì a + b < b + b

=> a + b < 2.b < a.b (vì a > 2)

Vậy với a > 2; b > 2 thì a + b < a.b (đpcm)

Ta có:

\(\frac{A}{B}=\frac{3}{7}\); \(\frac{B}{C}=\frac{15}{28}\)

=> \(\frac{A}{B}.\frac{B}{C}=\frac{3}{7}.\frac{15}{28}\)

=> \(\frac{A}{C}=\frac{45}{196}\)

Vậy tỉ số của 2 số C và A là \(\frac{196}{45}\)

Chọn A

\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

Đặt A = -B

\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2B=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(B=2-\frac{1}{1024}=\frac{2047}{1024}\)

=> \(A=-\frac{2047}{1024}\)