A=19^5+2015/19^5-1 và B=19^5+2014/19^5-2
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 A = 19^5+2015/19^5-1 ; B= 19^5+2014/19^5-2
\(A=\frac{19^5-1+2017}{19^5-1}\)
\(A=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{1+2017}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
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
LẤY (2015/19^5-1)-(2014/19^5-2)=(2015*19^5-2*2015-2014*19^5+2014)/((19^5-10*(19^5-2)
=(19^5-2016)/((19^5-1)*(19^5-2)>0
HAY A>B
Nhân tích chéo
mình chỉ gợi ý thôi bạn tự làm nhé
So sánh hai số sau: A =\(\frac{19^5+2016}{19^5-1}\) và B =\(\frac{19^5+2015}{^{19^5-2}}\)
\(A=\frac{19^5-1+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
ta thấy:B>1
=>\(B=\frac{19^5+2015}{19^5-2}>\frac{19^5+2015+1}{19^5-2+1}=\frac{19^5+2016}{19^5-1}=A\Rightarrow B>A\)
vậy.....
So sánh 2 số sau: \(A=\dfrac{19^5+2016}{19^5-1}\) và \(B=\dfrac{19^5+2015}{19^5-2}\)
Ta có: \(A=\frac{19^5+2016}{19^5-1}=\frac{19^5-1+2017}{19^5-1}=\frac{19^5-1}{19^5-1}+\frac{2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{19^5-2}{19^5-2}+\frac{2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(\frac{2017}{19^5-1}< \frac{2017}{19^5-2}\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\Rightarrow A< B\)
Vậy A < B
So sánh 2 số sau: \(A=\dfrac{19^5+2016}{19^5-1}\) và \(B=\dfrac{19^5+2015}{19^5-2}\)
\(A=\frac{19^5+2016}{19^5-1}=\frac{\left(19^5-1\right)+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{\left(19^5-2\right)+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
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 \(A< B\)
1 tính
a,( 17/20+19/21). 31/20 (17/20+19/21). (-31/20)
b, 3/4.15/2+15/8.(-3) + 199/200
c, (199/200+201/202+2015/2014). (1/2+2/3+ -7/6)
d, a(b+1/5) -a ( 6/5 + b ) với a= 2015/2015 ; b= 2000
So sánh 2 số sau: \(A=\dfrac{19^5+2016}{19^5-1}\) và \(B=\dfrac{19^5+2015}{19^5-2}\)
Ta có: \(A=\dfrac{19^5+2016}{19^5-1}=1+\dfrac{2017}{19^5-1}\)
\(B=\dfrac{19^5+2015}{19^5-2}=1+\dfrac{2017}{19^5-2}\)
Vì \(\dfrac{2017}{19^5-1}< \dfrac{2017}{19^5-2}\Rightarrow1+\dfrac{2017}{19^5-1}< 1+\dfrac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
Vậy A < B