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23 tháng 10 2017
\(A=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+...+\dfrac{1}{9^2.10}\)
\(A=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+....+\dfrac{1}{900}\)
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{3}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\........\\\dfrac{1}{900}< \dfrac{1}{32}\end{matrix}\right.\)
Nên \(A< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{32}.8=1\)
1 tháng 10 2017
\(LINH=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(LINH=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1^2.2^2}+\dfrac{2}{1^2.2^2}+\dfrac{3}{3^2.4^2}+\dfrac{4}{3^2.4^2}+\dfrac{5}{5^2.6^2}+\dfrac{6}{5^2.6^2}+\dfrac{7}{7^2.8^2}+\dfrac{8}{7^2.8^2}+\dfrac{9}{9^2.10^2}+\dfrac{10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+\dfrac{1}{7.8^2}+\dfrac{1}{7^2.8}+\dfrac{1}{9.10^2}+\dfrac{1}{9^2.10}\)\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+\dfrac{1}{448}+\dfrac{1}{392}+\dfrac{1}{900}+\dfrac{1}{810}\)Vì:
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{1}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\...............\\\dfrac{1}{810}< \dfrac{1}{32}\end{matrix}\right.\)
Nên:
\(\dfrac{1}{48}+\dfrac{1}{36}+.....+\dfrac{1}{810}< \dfrac{1}{32}+\dfrac{1}{32}+....+\dfrac{1}{32}\)
\(\Rightarrow\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{32}.8=\dfrac{1}{4}\)
Nên:
\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}=1\)
Nên \(LINH< 1\left(đpcm\right)\)
13 tháng 3 2020
a,\(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
=\(\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
=\(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)
= \(1-1=0\)
b,\(\left(\frac{-1}{3}\right)^2.\frac{4}{11}+\frac{7}{11}.\left(\frac{1}{3}\right)^2\)
=\(\frac{1}{9}.\frac{4}{11}+\frac{7}{11}.\frac{1}{9}\)
=\(\frac{1}{9}\left(\frac{4}{11}+\frac{7}{11}\right)\)
=\(\frac{1}{9}.1=\frac{1}{9}\)
c,\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
=\(\left(\frac{-2}{3}+\frac{3}{7}-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
=\(\left(-1+1\right).\frac{5}{4}=0.\frac{5}{4}\)=0
12 tháng 11 2017
a, (-0.25) * 7.9 * 40 = [ (-0.25) * 40] * 7.9 = -10 * 7.9 = -79
b, ( 3/2 )^3 * 2^3 = ( 3/2*2) ^3 = 3^3 = 27
c, (1.75 / \(\frac{7}{2}\)) * \(\frac{4}{5}\)= ( 7/4 * 7/2 ) *4/5 = \(\frac{49}{10}\)
d, \(\frac{11}{2}\cdot4\frac{5}{3}-2\frac{5}{3}\cdot\frac{11}{2}\)= \(\frac{11}{2}\cdot\left(\frac{17}{3}-\frac{11}{3}\right)\)= \(\frac{11}{2}\cdot\frac{6}{3}\)= \(\frac{11}{2}\cdot2\)=11
14 tháng 2 2020
\(C=...\)
\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)
\(=\frac{1}{3^2}-\frac{1}{15^2}\)
\(=\frac{1}{9}-\frac{1}{225}\)
Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)
\(=>C< \frac{1}{5}\)( ĐPCM )
14 tháng 2 2020
C = ...
=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)
C = \(\frac{1}{3^2}-\frac{1}{15^2}\)
Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)
=> C < \(\frac{1}{5}\)
NC
0
22 tháng 5 2022
Bài 1:
\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)
\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)
\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)
=1/131
Bài 2:
b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)
NC
23 tháng 2 2018
Mk lm đc câu nào hay câu đó nha 
A = 3/8 + [ 11/17 + (-3/8 ) ]
= [ 3/8 + ( -3 /8 ) ] + 11/17
= 0 + 11/17
= 11/17
C = 1/3 . 17/21 + 4/21 . 1/3 - 5/12
= 1/3 ( 17/21 + 4/21 ) - 5/12
= 1/3 - 5/12
= -1/12
D = 1/7 + 4/9 + 6/7 + 25/45
= ( 1/7 + 6/7 ) + ( 4/9 + 25/45 )
= 1 +1
= 2
Mk chỉ lm đc thế này thôi mong thông cảm 
hjhj
1/9 hoặc ra số thập phân 0,1111111111