So sánh:
A = \(\dfrac{5^{12}+1}{5^{13}+1}\) và B = \(\dfrac{5^{11}+1}{5^{12}+1}\)
K
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VN
Võ Ngọc Phương
CTVHS
VIP
10 tháng 10 2023
a) \(< \)
b) \(>\)
c) \(< \)
d) \(>\)
e) \(< \)
g) \(>\)
h) \(>\)
k) \(>\)
29 tháng 5 2022
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
19 tháng 2 2022
a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)
b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)
1 kg 
2 m
H
24 tháng 8 2023
a)
\(\dfrac{5}{8}\left(kg\right)< \dfrac{6}{8}\left(kg\right)\)
Do 13 > 12 nên \(\dfrac{13}{12}>1\)
hay: \(\dfrac{13}{12}\left(kg\right)>1\left(kg\right)\)
b)
\(\dfrac{11}{12}\left(l\right)>\dfrac{11}{14}\left(l\right)\)
Có:
\(\dfrac{5}{3}=\dfrac{5\times3}{3\times3}=\dfrac{15}{9}\)
nên: \(\dfrac{5}{3}\left(l\right)=\dfrac{15}{9}\left(l\right)\)
c)
\(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18n}\)
nên \(\dfrac{5}{6}\left(m\right)< \dfrac{17}{8}\left(m\right)\)
Có: \(2=\dfrac{2\times7}{7}=\dfrac{14}{7}\)
nên: \(\dfrac{16}{7}\left(m\right)>2\left(m\right)\)
O
2 tháng 4 2022
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
2 tháng 7 2023
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
KS
0
2 tháng 2
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
Áp dụng công thức \(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) \(\left(a;b;m\in N\right)\)
Ta có : \(A=\dfrac{5^{12}+1}{5^{13}+1}< 1\)
\(\Leftrightarrow A=\dfrac{5^{12}+1}{5^{13}+1}< \dfrac{5^{12}+1+4}{5^{13}+1+4}=\dfrac{5^{12}+5}{5^{13}+5}=\dfrac{5\left(5^{11}+1\right)}{5\left(5^{12}+1\right)}=B\)
\(\Leftrightarrow A< B\)