Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)

\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)

\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)

Vậy A<2

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

A; so sánh \(\frac{13^{15}+1}{13^{16}+1}\); \(\frac{13^{16}+1}{13^{17}+1}\)

\(\frac{13^{16}+1}{13^{17}+1}\) < \(\frac{13^6+\left(1+12\right)}{13^7+\left(1+12\right)}\) = \(\frac{13^{16}+13}{13^{17}+13}\) = \(\frac{13^{}.\left(13^{15}+1\right)}{13^{}.\left(13^{16}+1\right)}\)= \(\frac{13^{15}+1}{13^{16}+1}\)

Vậy \(\frac{13^{15}+1}{13^{16}+1}\)> \(\frac{13^{16}+1}{13^{17}+1}\)

Câu B:

\(\frac{1999^{2000}+1}{1999^{1999}+1}\) > \(\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\) = \(\frac{1999^{2000}+1999}{1999^{1999}+1999}\) = \(\frac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}\)

\(\frac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}\) = \(\frac{1999^{1999}+1}{1999^{1998}+1}\)

Vậy

\(\frac{1999^{1999}+1}{1999^{1998}+1}\) < \(\frac{1999^{2000}+1}{1999^{1999}+1}\)

Túi ko bt

Bạn xem lại đề câu a) cho rõ lại

Câu b) Tại x=2013 thì B=x²⁰¹³-(x+1)x²⁰¹²+(x+1)x²⁰¹¹-(x+1)x²⁰¹⁰+...-(x+1)x²+(x+1)x-1

                                 = x²⁰¹³-x²⁰¹³-x²⁰¹²+x²⁰¹²+x²⁰¹¹-x²⁰¹¹-x²⁰¹⁰+..-x³ - x²+x²+x-1

                                 = x-1 =  2012

phải là so sánh A với 2 mới đúng

a, A = \(\frac{1}{2}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)

\(A=\frac{1}{2}.\left(\frac{3.4....99}{4.5...100}\right)\)
\(A=\frac{1}{2}.\left(\frac{3}{100}\right)\)\(\)\(A=\frac{3}{200}\)

\(B=\frac{2}{3}.\frac{4}{5}.\frac{5}{6}...\frac{100}{101}\)

\(B=\frac{2}{3}.\left(\frac{4.5...100}{5.6...101}\right)\)

\(B=\frac{2}{3}.\left(\frac{4}{101}\right)\)

\(B=\frac{8}{303}\)

\(A.B=\frac{8}{303}.\frac{3}{200}\)

\(A.B=\frac{1}{2525}\)

b, A = 1/2 x 3/100

B = 2/3 x 4/101

Ta có : 1 - 2/3 = 1/3; 1 - 1/2 = 1/2

MÀ 1/3 < 1/2 => 2/3 > 1/2 (1)

Ta có : 1 - 3/100 = 97/100

1 - 4/101 = 97/101

Mà 97/101 < 97/100 => 4/101 > 3/100 (2)

Từ (1) và (2) => B > A

a,

\(AB=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)

\(AB=\frac{\left[1\cdot3\cdot5\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)

b,

1/2 < 2/3

3/4 < 4/5

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

99/100 < 100/101

=> \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\Leftrightarrow A< B\)

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

               \(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)

               \(=\frac{1.2....18.19}{2.3...19.20}\)

               \(=\frac{1}{20}>\frac{1}{21}\)

Vậy A > 1/21

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{98}{2^{98}}+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\) (lấy 2A - A = A)

Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

\(B=2B-B=2-\frac{1}{2^{99}}\)

Do đó: \(A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)

vế trước lớn hơn