1. TÍNH :
\(\frac{3.5+3.9.27+9.27.81+27.81.243}{4.5+3.12.27+9.36.81+27.108.243}\)
P/S : " . " là dấu nhân
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KN
6 tháng 5 2018
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
KN
7 tháng 5 2018
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
CT
0
8 tháng 3 2017
\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)
\(\Rightarrow S=\frac{2015}{672}\)
Vậy: \(\Rightarrow S=\frac{2015}{672}\)
Bạn giải giúp mk câu mk đăng tầm 5 phút nha!
12 tháng 5 2017
\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
30 tháng 6 2018
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{35.37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\frac{36}{37}\)
\(C=\frac{18}{37}\)
30 tháng 6 2018
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\frac{36}{37}=\frac{18}{37}\)
Vay C = \(\frac{18}{37}\)
HT
0
MH
7 tháng 1 2016
\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
=> 2S + 1/101 = \(2.\frac{50}{101}+\frac{1}{101}=\frac{100}{101}+\frac{1}{101}=\frac{101}{101}=1\)
Gọi P/s trên là S, ta có :
\(S=\frac{3.5+3.9.27+9.27.81+27.81.243}{4.5+3.12.27+9.36.81+27.108.243}\)
\(S.\frac{1}{3}=\frac{\left(1\right).5+3.\left(3\right).27+9.\left(9\right).81+27.\left(27\right).243}{4.5+3.12.27+9.36.81+27.108.243}\)
\(S.\frac{1}{3}=\frac{1}{4}\)
\(\Rightarrow\)\(S=\frac{1}{4}:\frac{1}{3}=\frac{3}{4}\)