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Ta có:
1-45/46 = 1/46
1-98/99 = 1/99
Vì 1/46 > 1/99
=> 45/46 > 98/99
Chúc bạn học tốt!
Ta có:
\(\frac{45}{46}=1-\frac{1}{46}\)
\(\frac{98}{99}=1-\frac{1}{99}\)
Vì \(\frac{1}{46}>\frac{1}{99}\) nên \(\frac{46}{46}< \frac{98}{99}\)
Ta có :
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=50.\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow\)\(S>\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\)
Chúc bạn học tốt ~
Bài 1:
Ta thấy A < 1
=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)
Vậy A < B
Bài 2:
Ta thấy C < 1
=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C < D
\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)
\(B=\frac{98^{99}+1}{98^{89}+1}\)
A-1=\(\frac{98^{98}-98^{88}}{98^{88}+1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}\)
B-1=\(\frac{98^{99}-98^{89}}{98^{89}+1}=\frac{98^{89}.\left(98^{10}-1\right)}{98^{89}+1}\)
=>\(\frac{A-1}{B-1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}.\frac{98^{89}+1}{98^{89}.\left(98^{10}-1\right)}=\frac{98^{89}+1}{98.\left(98^{88}+1\right)}=\frac{98^{89}+1}{98^{89}+98}< 1\)
->A-1<B-1
->A<B
a ) Ta có :
\(1-\frac{41}{91}=\frac{50}{91}\) \(=\frac{500}{910}\) ; \(1-\frac{411}{911}=\frac{500}{911}\)
Vì \(\frac{500}{910}>\frac{500}{911}\)nên \(\frac{41}{91}< \frac{411}{911}\)
b ) Ta có :
\(1-\frac{113}{115}=\frac{2}{115}\) ; \(1-\frac{93}{95}=\frac{2}{95}\)
Vì \(\frac{2}{115}< \frac{2}{95}\)nên \(\frac{113}{115}>\frac{93}{95}\).
c ) Quy đồng TS ta có :
\(\frac{13}{53}=\frac{143}{583}\) ; \(\frac{11}{30}=\frac{143}{390}\)
Vì \(\frac{143}{583}< \frac{143}{390}\)nên \(\frac{13}{53}< \frac{11}{30}\).
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{17}+1}{17^{18}+1}=B\)
=> A < B