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Ta có :
\(VT=\frac{1}{2}\left[\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)^2}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\frac{\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2+\left(a-c\right)^2+\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{a^2+b^2+c^2-ab-bc-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)(1)
Lại có :
\(VP=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{\left(b-c\right)\left(a-c\right)+\left(a-b\right)\left(a-c\right)-\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab-bc-ac+c^2+a^2-ac-ab+bc-ab+ac+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2+b^2+c^2-ab-ac-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)(2)
Từ (1) và (2) \(\RightarrowĐPCM\)
a) \(\frac{1}{a\left(a+1\right)}=\frac{a+1-a}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)
b)\(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
Sai rồi thê này nè
a/ \(\frac{1}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)
Ta co: \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)
b/ \(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta co: \(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}\)
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
