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Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
#)Giải :
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)
\(\Rightarrow5A=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\)
\(\Rightarrow5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)
\(\Rightarrow5A=\frac{1}{4}-\frac{1}{49}=\frac{45}{196}\)
\(\Rightarrow A=\frac{45}{196}\div5=\frac{9}{196}\)
Thay A vào B, ta được :
\(B=\frac{9}{196}.\frac{1-3-5-...-49}{89}\)
\(B=\frac{9}{196}.\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(B=\frac{9}{196}.\frac{1-\left[\frac{\left(49+3\right).\left(\frac{49-3}{2}+1\right)}{2}\right]}{89}\)
\(B=\frac{9}{196}.\frac{-623}{89}=-\frac{9}{28}\)
B = \(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)
= \(\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\right).\frac{1-\left(3+5+7+...+49\right)}{89}\)
= \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(24.52:2\right)}{89}\)
= \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-624}{89}\)
= \(\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)
= \(\frac{-9}{28}\)
Vậy B = \(-\frac{9}{28}\)
A= \(\frac{\left(5^4-5^3\right)^3}{125^3}=\frac{\left(625-125\right)^3}{1953125}=\frac{125000000}{1953125}=64\)
B=\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{729}{2916}=\frac{1}{4}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{9}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{8}{9}\)
\(A=\frac{1}{9}\)
\(\Rightarrow\)A= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}\frac{8}{9}\)
\(\Rightarrow\)A=\(\frac{1.2.3.4.5.6.7.8}{2.3.4.5.6.7.8.9}\)
\(\Rightarrow\)A=\(\frac{1}{9}\)
HỌC TỐT!!!
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
0==)=====>
\(M=-\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{\left(n+4\right).n}\right)\)
\(M=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)
\(M=-\left(1-\frac{1}{n+4}\right)\)
\(M=-1+\frac{1}{n+4}=\frac{-n-3}{n+4}\)