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Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)
= \(\frac{1}{2}.\frac{390}{781}=\frac{195}{781}\)
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{998.999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{998.999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{998.999}-\frac{1}{999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{999.1000}=\frac{499499}{999000}\Leftrightarrow A=\frac{499499}{1998000}\)
\(B=\frac{1}{1.2.3.4.5}+\frac{1}{2.3.4.5.6}+\frac{1}{3.4.5.6.7}+\frac{1}{996.997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{4}{1.2.3.4.5}+\frac{4}{2.3.4.5.6}+\frac{4}{3.4.5.6.7}+....+\frac{4}{996.997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{2.3.4.5}+\frac{1}{2.3.4.5}-\frac{1}{3.4.5.6}+\frac{1}{3.4.5.6}-\frac{1}{4.5.6.7}+...+\frac{1}{996.997.998.999}-\frac{1}{997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{997.998.999.1000}=\frac{41417124749}{994010994000}\Leftrightarrow B=\frac{41417124749}{3976043976000}\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2015}\right)\times\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2014}{2015}\times\frac{2015}{2016}\)
\(=\frac{1}{2016}\)
\(=\frac{1}{2}.\frac{2}{3}...\frac{2015}{2016}=\frac{1.2....2015}{2.3....2016}=\frac{1}{2016}\)
Bạn xem lại đề, là cộng mới đúng chứ ???
Mình làm được rồi này :
\(B=\frac{1}{1.2.3}-\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{97.98.99}\right)\)
\(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{97.98}-\frac{1}{98.99}\right)\)
\(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{98.99}\right)\)
\(=\frac{1}{6}-\frac{1}{6}+\frac{1}{9702}\)
\(=\frac{1}{9702}\)