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Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
C= 101 . 35 . 10101.23 / 35.10101 . 23. 101 = 1
Do D = 3535/3534 > 1 => C<D
a, 135 / 120 < 13/8
b, 3535/4848 > 5/8
c, 650650/480480 < 222222/144144
Tk cho mk nha!!!!
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
\(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}+\frac{2}{10\times12}+\frac{2}{12\times14}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}+\frac{1}{10}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}=\frac{3}{7}\)
a) \(\frac{7}{9}\)< 1 còn \(\frac{9}{7}\)> 1
Vậy \(\frac{7}{9}< \frac{9}{7}\)
b) Ta rút gọn phân số \(\frac{3535}{4848}\)= \(\frac{35}{48}\)
\(\frac{35}{48}\)và \(\frac{5}{8}\)MSC: 48
Ta có:
Giữ nguyên phân số \(\frac{35}{48}\) \(\frac{5}{8}=\frac{5x6}{8x6}=\frac{30}{48}\)
Vì \(\frac{35}{48}>\frac{30}{48}\)nên \(\frac{3535}{4848}>\frac{5}{8}\)
a) ta có : 7/9 <1 và 9/7 > 1 => 7/9 < 9/7
b) 3535/4848 = 35/48
ta có 5/ 8 = 30/48
=> 3535/4848> 5/8
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9
2A=2/1.3+2/3.5+2/5.7+2/7.9
2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
2A=1-1/9
2A=8/9
A=4/9
a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
\(\frac{3636}{3535}+\frac{5454}{4242}\)
\(=\frac{36}{35}+\frac{54}{42}\)
= \(\frac{36}{35}+\frac{9}{7}\)
= \(\frac{36}{35}+\frac{45}{35}\)
= \(\frac{81}{35}\)
81/35