1 + 2 + 2² + 2³ + ... + 2^\(x\) = 127

Đặt vế trái bằng: A ta có:

  A = 1 + 2 + 2² + ... + 2^\(x\) 

2A = 2 + 2² + 2³ +...+ 2^{\(^{x+1}\)}

2A - A = (2 + 2² + 2³ +... + 2^{\(^{x+1}\)} - (1 + 2 + 2² + ...+ 2^{\(^x\)})

A         = 2 + 2² + 2\(^3\) +..+ 2\(^{x+1}\) - 1 - 2 - 2² - ... - 2^{\(^x\)} 

A         = (2 - 2) + (2² - 2²) + ... +(2\(^x\) - 2\(^x\)) + (2\(^{x-1}\) - 1)

A         =  0 + 0 + ... + 0  + 2^{\(^{x-1}\)} - 1

A        = 2^{\(^{x+1}\)}  - 1

2^{\(^{x+1}\)} - 1 = 127

2\(^{x+1^{ }}\)      = 127 + 1

2\(^{x+1}\)      = 128

2\(^{x+1}\)      = 7⁷

  \(x\) + 1    = 7

   \(x\)          = 7  - 1

    \(x\)        = 6

\(x=6.\)

Vì : 4/5 > 6/11 > 5/9 nên trong 3 ngày ngày thứ nhất bán vải màu hồng là được nhiều nhất

Thầy ơi, vải tím là \(\dfrac{7}{11}\) chứ không phải \(\dfrac{6}{11}\) nhé thầy.

\(a,MSC:360\\ -\dfrac{3}{8}=\dfrac{-3.45}{8.45}=\dfrac{-135}{360};\dfrac{11}{24}=\dfrac{11.15}{24.15}=\dfrac{165}{360};\dfrac{11}{18}=\dfrac{11.20}{18.20}=\dfrac{220}{360}\\ ;\dfrac{-7}{9}=\dfrac{-7.40}{9.40}=\dfrac{-280}{360};\dfrac{4}{5}=\dfrac{4.72}{5.72}=\dfrac{288}{360}\\ b,MSC:135\\ \dfrac{-4}{15}=\dfrac{-4.9}{15.9}=\dfrac{-36}{135};\dfrac{7}{9}=\dfrac{7.15}{9.15}=\dfrac{105}{135};\dfrac{-1}{3}=\dfrac{-1.45}{3.45}=\dfrac{-45}{135}\\ ;\dfrac{2}{5}=\dfrac{2.27}{5.27}=\dfrac{54}{135};0=\dfrac{0}{135};\dfrac{13}{45}=\dfrac{13.3}{45.3}=\dfrac{39}{135}\)

ummm..khó quá

\(a,\dfrac{2}{3}< \dfrac{x}{12}< \dfrac{7}{6}\\ \dfrac{2.4}{3.4}< \dfrac{x}{12}< \dfrac{7.2}{6.2}\\ \dfrac{8}{12}< \dfrac{x}{12}< \dfrac{14}{12}\\ Vậy:x\in\left\{9;10;11;12;13\right\}\\ PS:\left\{\dfrac{9}{12};\dfrac{10}{12};\dfrac{11}{12};\dfrac{12}{12};\dfrac{13}{12}\right\}\)

\(b,\dfrac{25}{9}< \dfrac{50}{x}< \dfrac{10}{3}\\ \dfrac{25.2}{9.2}< \dfrac{50}{x}< \dfrac{10.5}{3.5}\\ \dfrac{50}{18}< \dfrac{50}{x}< \dfrac{50}{15}\\ Nên:x\in\left\{17;16\right\}\\ Vậy:PS\left\{\dfrac{50}{17};\dfrac{50}{16}\right\}\)

a; 8 = 2³; 24 = 2³.3; 18 = 2.3²; 5 = 5

 BCNN(8; 24; 18; 5) = 2³.3².5 = 360

\(\dfrac{-3}{8}\) = \(\dfrac{-3.45}{8.45}\) = \(\dfrac{-135}{360}\);  \(\dfrac{11}{24}\) =  \(\dfrac{11.15}{24.15}\) = \(\dfrac{165}{360}\) 

\(\dfrac{11}{18}\) = \(\dfrac{11.20}{18.20}\) = \(\dfrac{220}{360}\); \(\dfrac{-7}{9}\) = \(\dfrac{-7.40}{9.40}\) = \(\dfrac{-280}{360}\)

 \(\dfrac{4}{5}\) = \(\dfrac{4.72}{5.72}\) = \(\dfrac{288}{360}\); 

\(\dfrac{-7}{9}\); \(\dfrac{-3}{8}\); \(\dfrac{11}{24}\); \(\dfrac{11}{18}\); \(\dfrac{4}{5}\)

b; 15 = 3.5; 9 = 3²; 3 = 3; 45 = 3².5

BCNN(15; 9; 45) = 3².5

\(\dfrac{-4}{15}\) = \(\dfrac{-4.3}{15.3}\) = \(\dfrac{-12}{45}\); \(\dfrac{7}{9}\) = \(\dfrac{7.5}{9.5}\) = \(\dfrac{35}{45}\); \(\dfrac{-1}{3}\) = \(\dfrac{-1.15}{3.15}\) = \(\dfrac{-15}{45}\)

\(\dfrac{2}{5}\) = \(\dfrac{2.9}{5.9}\) =  \(\dfrac{18}{45}\); 0 = \(\dfrac{0}{45}\); \(\dfrac{13}{45}\)

\(\dfrac{-1}{3}\); \(\dfrac{-4}{15}\); 0; \(\dfrac{13}{45}\); \(\dfrac{2}{5}\);  \(\dfrac{7}{9}\)

\(A=2^0+2^17+2^2+..+2^{19}\)

\(2A=2+2^2+...+2^{20}\)

\(2A-A=2+2^2+...+2^{20}-1-2-...-2^{19}\)

\(A=2^{20}-1\)

\(7.\left(-23\right)+7.23\)

\(=7.23\left(-1+1\right)\)

\(=7.23.0=0\)

7.(-23)+7.23

= -7. 23 + 7.23

= 23.7 - 23.7

= 0

Gọi d=UCLN(a;b)

=> Tồn tại 2 số nguyên m;n sao cho

a=md và b=nd

ta có

a+b=md+nd=d(m+n)=p\(\Rightarrow p⋮d\) mà p là số nguyên tố nên d=1

=> a và b nguyên tố cùng nhau

\(D=\dfrac{6-1}{1.6}+\dfrac{11-6}{6x11}+\dfrac{16-11}{11x16}+...+\dfrac{56-51}{51.56}=\)

\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{51}-\dfrac{1}{56}=1-\dfrac{1}{56}=\dfrac{55}{56}\)