THEO MINH LA VAY NE:

A<1/3

mình cũng vậy

<1/3 k nhé:)

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

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

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

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

A < 1 - \(\frac{1.}{100}\)

A < \(\frac{99}{100}< \frac{199}{100}\)

=> A < \(\frac{199}{100}\)

b,

S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)

S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)

S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)

S = \(\frac{1.11}{2.10}\)

S = \(\frac{11}{20}\)

A =\(\frac{\left(\frac{17}{5}+\frac{1}{5}\right).\frac{2}{5}}{\left(\frac{38}{7}-\frac{9}{4}\right).\frac{56}{267}}\)

A=\(\frac{36}{25}\).\(\frac{3}{2}\)=\(\frac{54}{25}\)=2,16

B=\(\frac{1,2:\left(\frac{6}{5}-\frac{5}{4}\right)}{0,32+\frac{2}{25}}\)=-24.\(\frac{5}{2}\)=-60

vì 2,16 > -60 Vậy A>B

Ta có

\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)                                                   \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)

\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\)                                            \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)

\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\)                                                            \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)

\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\)                                                                       \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)

\(\Leftrightarrow A=\frac{4968}{2225}\)                                                                      \(\Leftrightarrow B=\frac{25}{8}\)

\(\Leftrightarrow A=\frac{39744}{17800}\)                                                                     \(\Leftrightarrow B=\frac{55625}{17800}\)

Ta có: 39744<55625

\(\Rightarrow A< B\)

Vậy A<B

kb vói mình đã

1) Đặt dãy trên là \(A\)

Theo bài ra ta có :

\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )

\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)

\(\Rightarrow A< B\)

Vậy \(A< B.\)

1) Ta có B =

 \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)

=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)

Sai thì thôi chứ mk chỉ làm rờ thôi

1) Ta có :\(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

\(\RightarrowĐPCM\)

1/100 hả e hay là 1/10

Dạ 1/100

theo đây mà làm giờ a bận chuẩn bị KT 1T r

link:  https://olm.vn/hoi-dap/question/148148.html