Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
So sanh
\(\frac{1}{\left(-2008\right)^{2008}}\) va \(\text{}\text{}\frac{1}{\left(-2009\right)^{2009}}\)
BAi 1
a) hay so sanh
\(\frac{2014}{2015}+\frac{2015}{2014}\)va \(\frac{666665}{333333}\)
b)tinh
A=\(\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)\left(1-\frac{1}{45}\right)...\left(1-\frac{1}{1326}\right)\)
cac ban lam nhanh nhe minh dang can gap xong minh tick cho
so sanh :
\(\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{9}\right)-\left(1-\frac{1}{16}\right)-......-\left(1-\frac{1}{10000}\right)\) voi\(\frac{1}{2}\)
so sanh
a, \(\left|-2\right|^{300}\) va \(\left|-4\right|^{150}\)
b,\(\left|-2\right|^{300}\)va \(\left|-2\right|^{300}\)
co loi giai
cho so A=\(\frac{2013+\frac{1}{2}}{\left(2012+\frac{1}{2}\right)^2+2013+\frac{1}{2}}\)
B=\(\frac{2013+\frac{1}{3}}{\left(2012+\frac{1}{3}\right)^2+2013+\frac{1}{3}}\)
so sanh A va B
Cho A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\)
CMR:A=\(\frac{1}{299}.\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.....+\frac{1}{400}\right)\)
ở chỗ 1/299 là nhân với ngoặc vuông nha bạn nào giải hộ mình rthì i li-ke
so sanh
\(\left(\frac{1}{243}\right)^9va\left(\frac{1}{83}\right)^{13}\)