CMR:
A) \(\frac{113}{342}>\frac{251}{755}\) VÀ
B)\(\frac{1784}{3456}>\frac{673}{1234}\) là ko đúng
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cu lay phep tinh nay tru phep tinh kia hk ra thi nt hoi mink
gọi 1/41+1/42+1/43+...+1/79+1/80 là A
ta có:1/41>1/60,1/42>1/60,1/43>1/60,...,1/60=1/60
=>1/41+1/42+1/43+...+1/60>1/60
1/61>1/80,..................................,1/80=1/80
=>1/61+1/62+............+1/80>1/80
=>1/41+1/42+1/43+...+1/79+1/80>1/60+1/80
lại có 7/12=1/4+1/3
1/60.20=1/3 và 1/80.20=1/4
=>1/41+1/42+1/43+...+1/79+1/80>1/3+1/4
=>1/41+1/42+1/43+...+1/79+1/80>7/12
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)
\(=\frac{1}{7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\frac{36}{37}\)
\(=\frac{6}{37}\)
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)
\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{35.37}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
~ Hok tốt ~
Minh chi biet lam cau b thoi ak
b) Giai:
B=10^16+1 tren 10^17 +1 <10^16+1+9 tren 10^17+1+9
ma 10^16+1+9 tren 10^17+1+9 = 10^16+10 tren 10^17+10
=10(10^15+1) tren 10(10^16+1)
=10^15+1 tren 10^16+1 =A
=>A>B
Cho y kien voi!
http://olm.vn/hoi-dap/question/126681.html
Bạn tham khảo nhé
\(A=\frac{79}{1999}+\frac{191}{1998}+\frac{947}{1997}+\frac{673}{1998}+\frac{110}{1999}\)
\(A=\left(\frac{79}{1999}+\frac{110}{1999}\right)+\left(\frac{191}{1998}+\frac{673}{1998}\right)+\frac{947}{1997}\)
\(A=\frac{189}{1999}+\frac{16}{37}+\frac{947}{1997}\)
a
sai rồi B