So sánh các phân số sau
c)\(\frac{11}{15}\)và\(\frac{13}{17}\)
d)\(\frac{18}{53}\)và\(\frac{26}{78}\)
e)\(\frac{72}{73}\)và\(\frac{98}{99}\)
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a, \(\frac{-11}{12}>\frac{17}{-18}\)
b\(\frac{2}{5}< \frac{5}{7}\)
c\(\frac{-3}{4}>\frac{-6}{7}\)
d\(\frac{19}{18}>\frac{2005}{2004}\)
e\(\frac{72}{73}< \frac{98}{99}\)
a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b) \(\frac{99}{100}\)< \(\frac{100}{99}\); c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\);
d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)
e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)
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
13/21<9/11
6/17<9/25
7/12>11/-18
tk mk nha mk đang âm điểm
chúc các bn hok tốt ^-^
Ta co : 72/73 = 1 - 1/73 ; 98/99 = 1 - 1/99
Vì 1/73 > 1/99 suy ra 1 - 1/73 < 1 - 1/99 hay 72/73 < 98/99
Ta có: \(\frac{72}{73}=1-\frac{1}{73}\);\(\frac{98}{99}=1-\frac{1}{99}\)
Vì \(\frac{1}{73}>\frac{1}{99}\Rightarrow1-\frac{1}{73}< 1-\frac{1}{99}\)
\(\Rightarrow\frac{72}{73}< \frac{98}{99}\)
Vậy \(\frac{72}{73}< \frac{98}{99}\)