\(A=\frac{19^5+2016}{19^5-1}=\frac{\left(19^5-1\right)+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{\left(19^5-2\right)+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(19^5-1>19^5-2\) nên \(\frac{2017}{19^5-1}< \frac{2}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
Vậy \(A< B\)
So sánh 2 số sau: \(A=\dfrac{19^5+2016}{19^5-1}\) và \(B=\dfrac{19^5+2015}{19^5-2}\)
So sánh hai số sau: A =\(\frac{19^5+2016}{19^5-1}\) và B =\(\frac{19^5+2015}{^{19^5-2}}\)
So sánh 2 số sau: A=19^5+2015/19^5-1 và B=19^5+2014/19^5-2
So sánh hai số sau A=19^5+2015/19^5-1. B=19^5+2014/19^5-2
So sánh 2 số sau
A= 19^5+2015/19^5-1 và 19^5+2014/19^5-2
Giải nhanh giúp mình nha. Thank
So sánh A = 19^5+2015/19^5-1 ; B= 19^5+2014/19^5-2
so sánh:
A=195+2016:195-1 với B=195+2015:195-2
Tính bằng cách hợp lý:
\(a,\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)\)
\(b,\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)
\(c,\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(d,6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(e,\left(31,12-5,97\right)-\left(-68,88+4,03\right)\)
\(h,3,7.\left(-10,56\right)+3,7.110,56\)
cho A =\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) so sánh A với 1
