Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)
B=\(\frac{3}{4}\)
So sánh A và B
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 4 2017
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 = \(\frac{11}{20}\)
NN
5 tháng 2 2016
Ta có:
\(2^2<4^2\Rightarrow\frac{1}{2^2}>\frac{1}{4^2}\)
\(3^2<6^2\Rightarrow\frac{1}{3^2}>\frac{1}{6^2}\)
\(4^2<8^2\Rightarrow\frac{1}{4^2}<\frac{1}{8^2}\)
\(...\)
\(100^2<200^2\Rightarrow\frac{1}{100^2}>\frac{1}{200^2}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
\(\Rightarrow A>B\)
9 tháng 7 2017
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
9 tháng 7 2017
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
=> \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\Leftrightarrow A< B\)
24 tháng 4 2017
a/ Ta có
\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)
\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
Thế lại bài toán ta được:
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)
24 tháng 4 2017
b/ Ta có:
A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)
\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)
Vậy A < B
WH
1 tháng 5 2018
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
NT
Cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
so sánh B với \(\frac{3}{4}\)
1
LA
27 tháng 3 2018
Ta có:\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)
B < \(\frac{1}{4}\) < \(\frac{3}{4}\)
\(\Leftrightarrow B< \frac{3}{4}\)
NH
Cho \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}};B=\frac{1}{2}\).so sánh A và B
1
AH
Akai Haruma
Giáo viên
6 tháng 7
Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}$
$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$
$\Rightarrow 3A-A=1-\frac{1}{3^{100}}$
$\Rightarrow 2A=1-\frac{1}{3^{100}}<1$
$\Rightarrow A< \frac{1}{2}$
$\Rightarrow A< B$
hi hi dễ ẹt
A<B