Bài 1: so sánh
a)\(\frac{14}{21}va\frac{60}{72}\)
b)\(\frac{38}{133}va\frac{129}{344}\)
c)\(\frac{11}{54}va\frac{22}{37}\)
d) \(A=\frac{10^{1990}+1}{10^{1991}+1}vaB=\frac{10^{1991}+1}{10^{1992}+1}\)
K
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CP
4 tháng 6 2020
a) \(\frac{2}{3}=\frac{8}{12}\) ; \(\frac{1}{4}=\frac{3}{12}\)
mà 8 > 3 ⇒ \(\frac{8}{12}>\frac{3}{12}\)⇒\(\frac{2}{3}>\frac{1}{4}\)
b) \(\frac{7}{10}\) và \(\frac{7}{8}\); mà 10 > 8 ⇒ \(\frac{7}{10}< \frac{7}{8}\)
c) \(\frac{6}{7}=\frac{30}{35}\); \(\frac{3}{5}=\frac{21}{35}\)
mà 30 > 21 ⇒ \(\frac{30}{35}>\frac{21}{35}\)⇒\(\frac{6}{7}>\frac{3}{5}\)
d) \(\frac{14}{21}=\frac{2}{3}\); \(\frac{60}{72}=\frac{5}{6}\)
\(\frac{2}{3}=\frac{4}{6}\) ⇒ \(\frac{2}{3}< \frac{5}{6}\)⇒ \(\frac{14}{21}< \frac{60}{72}\)
e) \(\frac{38}{133}=\frac{2}{7}\); \(\frac{129}{344}=\frac{3}{8}\)
\(\frac{2}{7}=\frac{16}{56}\) ; \(\frac{3}{8}=\frac{21}{56}\) mà 16<21 ⇒ \(\frac{16}{56}< \frac{21}{56}\)⇒ \(\frac{38}{133}< \frac{129}{344}\)
f) \(\frac{11}{54}=\frac{22}{108}\)và \(\frac{22}{37}\) mà 108 > 37 ⇒ \(\frac{22}{108}< \frac{22}{37}\)⇒ \(\frac{11}{54}< \frac{22}{37}\)
SD
21 tháng 4 2019
dấu "/" là phân số nha
VD: 2/3 là \(\frac{2}{3}\)do mk lười gõ wá
giúp mk giải
!
21 tháng 4 2019
1) So sánh :
\(\frac{2}{3}\)và \(\frac{1}{4}\)
\(\frac{2}{3}\)=\(\frac{2.4}{3.4}\)=\(\frac{8}{12}\); \(\frac{1}{4}\)=\(\frac{1.3}{4.3}\)=\(\frac{3}{12}\)
Vì \(\frac{8}{12}>\frac{3}{12}\)(8>3 )
Nên \(\frac{2}{3}>\frac{1}{4}\)
TT
2
3 tháng 5 2015
\(A=\frac{10^{1990}+1}{10^{1991}+1}vàB=\frac{10^{1991}+1}{10^{1992}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}
BD
10 tháng 5 2017
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+1+9}{10^{1991}+1}\Rightarrow10A=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+1+9}{10^{1992}+1}\Rightarrow10B=1+\frac{9}{10^{1992}+1}\)
=> 10A > 10B
=> A>B
LT
3
18 tháng 7 2016
Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
=> \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
=> \(B< \frac{10^{1991}+10}{10^{1992}+10}\)
=> \(B< \frac{10.\left(10^{1990}+1\right)}{10.\left(10^{1991}+1\right)}\)
=> \(B< \frac{10^{1990}+1}{10^{1991}+1}=A\)
=> B < A
VK
4
20 tháng 4 2017
Ta có : \(A=\frac{10^{1990}+1}{10^{1991}+1}=>10A=\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=>10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{\left(10^{1991}+1\right)+9}{10^{1991}+1}\)
\(=>10A=1+\frac{9}{10^{1991}+1}\)
Ta lại có : \(B=\frac{10^{1991}+1}{10^{1992}+1}=>10B=\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
Tương tự như A => \(10B=1+\frac{9}{10^{1992}+1}\)
Vì \(\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}=>10A>10B\)
\(=>A>B\)
TB
3
11 tháng 5 2017
Ta có :
A = \(\frac{10^{1990}+1}{10^{1991}+1}\)
10A = \(\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+10}{10^{1991}+1}\)
10A = \(\frac{10^{1991}+1+9}{10^{1991}+1}\)
10A = \(1+\frac{9}{10^{1991}+1}\left(1\right)\)
Ta lại có :
B = \(\frac{10^{1991}+1}{10^{1992}+1}\)
10B = \(\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+10}{10^{1992}+1}\)
10B = \(\frac{10^{1992}+1+9}{10^{1992}+1}\)
10B = \(1+\frac{9}{10^{1992}+1}\left(2\right)\)
Từ \(\left(1\right)va\left(2\right)\)
Ta có :\(1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\)10A > 10B
\(\Rightarrow\)A > B
TN
3
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
31 tháng 7 2019
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Vì \(10^{1991}< 10^{1992}\Rightarrow1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\frac{10^{1990}+1}{10^{1991}+1}>\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow A>B\)
31 tháng 7 2019
Ta có : \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Mà : \(\frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}\)
\(=\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)
\(=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow B< A\)
VT
2
2 tháng 5 2015
Ta có:
\(A=\left(\frac{10^{1990}+1}{10^{1991}+1}\right).\frac{10}{10}=\frac{10^{1991}+10}{10^{1992}+10}\)
Mình làm bằng cách tính phần bù:
Ta có:
\(1-A=1-\frac{10^{1991}+10}{10^{1992}+10}=\frac{10^{1992}+10}{10^{1992}+10}-\frac{10^{1991}+10}{10^{1992}+10}=\frac{10^{1992}-10^{1991}}{10^{1992}+10}\)
\(1-B=1-\frac{10^{1991}+1}{10^{1992}+1}=\frac{10^{1992}+1}{10^{1992}+1}-\frac{10^{1991}+1}{10^{1992}+1}=\frac{10^{1992}-10^{1991}}{10^{1992}+1}\)
Vì \(\frac{10^{1992}-10^{1991}}{10^{1992}+10}\frac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow A>B\)
2 tháng 5 2015
Vì\(\frac{10^{1991}+1}{10^{1992}+1}\)<1
Nên\(\frac{10^{1991}+1}{10^{1992}+1}\)<\(\frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Ta có: \(\frac{10^{1991}+1+9}{10^{1992}+1+9}\)=\(\frac{10^{1991}+10}{10^{1992}+10}\)=\(\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\frac{10^{1990}+1}{10^{1991}+1}\)
=>\(\frac{10^{1991}+1}{10^{1992}+1}\)<\(\frac{10^{1990}+1}{10^{1991}+1}\)
Vậy: B<A
KS
2
25 tháng 4 2019
Đặt \(A=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow10A=\frac{10\cdot(10^{1990}+1)}{10^{1991}+1}\)
\(=\frac{10^{1991}+10}{10^{1991}+1}=\frac{10^{1991}+1+9}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
Đặt \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow10B=\frac{10\cdot(10^{1991}+1)}{10^{1992}+1}=\frac{10^{1992}+10}{10^{1992}+1}=\frac{10^{1992}+1+9}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Tự so sánh được rồi -_-
a) \(\frac{14}{21}=\frac{2}{3}=\frac{4}{6}\)
\(\frac{60}{72}=\frac{5}{6}\)
Vì \(\frac{4}{6}< \frac{5}{6}\)
nên \(\frac{4}{21}< \frac{60}{72}\)