14- ....+3=10
dien so
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TT
0
TT
0
TT
2
SY
1 tháng 4 2016
Ta có: A=\(\frac{14^{15}+3}{14^{15}+3}\) = 1
B=\(\frac{14^{16}+5}{14^{17}+5}\) < 1 => B<1=A => B<A.
15 tháng 9 2017
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2 tháng 7 2016
Co ai tra loi duoc ko ????
ko ai tra loi duoc u??? Cac ban lop 9,8,7 cung ko lam duoc a
ko lam duoc mk lam cho nhung k nha
2 tháng 7 2016
Neu bot o so do 9 don vi ta duoc so moi chia het cho 11 va 14. muon thuong giua so moi va 11 cung bang thuong giua so moi voi 1 khi chia so moi cho 11 can bot o tu so moi la : 11 x 3 = 333 [don vi]
hieu giua 14 va 11 la
14 - 11 = 3
thuong giua so ben phai tim voi 14 la
33 : 3 = 11
so phai tim la
14 x 11 + 9 = 163
k nha
MT
4
MB
1
28 tháng 10 2021
\(1,2\sqrt{3}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}=\sqrt{75}>\sqrt{74}\\ 2,\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\\ \left(2\sqrt{2}+\sqrt{6}\right)^2=14+4\sqrt{3}\\ 6\sqrt{5}=\sqrt{180}>\sqrt{48}=4\sqrt{3}\\ \Rightarrow3+\sqrt{5}>2\sqrt{2}+\sqrt{6}\\ 3,3\sqrt{3}+\sqrt{26}+1>3\sqrt{3}+\sqrt{3}\left(\sqrt{26}>\sqrt{3}\right)\\ \Rightarrow\sqrt{27}+\sqrt{26}+1>4\sqrt{3}=\sqrt{48}\\ 4,\dfrac{1}{\sqrt{15}+\sqrt{14}}< \dfrac{1}{\sqrt{14}+\sqrt{13}}\left(\sqrt{15}+\sqrt{14}>\sqrt{14}+\sqrt{13}\right)\\ \Rightarrow\sqrt{15}-\sqrt{14}< \sqrt{14}-\sqrt{13}\)
TT
1
27 tháng 1 2022
\(14A=\dfrac{14^{16}+42}{14^{16}+3}=1+\dfrac{39}{14^{16}+3}\)
\(14B=\dfrac{14^{17}+70}{14^{17}+5}=1+\dfrac{65}{14^{17}+5}\)
mà \(\dfrac{39}{14^{16}+3}< \dfrac{65}{14^{17}+5}\)
nên 14A<14B
hay A<B
14-7+3=10
so 7
k mik nha mik k lai
14 - 7 + 3 = 10