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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,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
a) Ta có: \(A=\frac{3535.232323}{353535.2323}=\frac{35.101.23.10101}{35.10101.23.101}=\frac{35}{23}\)
\(B=\frac{3535}{3434}=\frac{35.101}{34.101}=\frac{35}{34}\)
\(C=\frac{2323}{2322}=\frac{2323}{2322}\)
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))
= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))
= 12 x \(\frac{1}{2}\) x ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
= 6 x ( \(\frac{1}{3}\) - \(\frac{1}{11}\))
= 6 x \(\frac{8}{33}\)
= \(\frac{48}{33}\)=\(\frac{16}{11}\)
Nhớ tk nha
\(\frac{3636}{3535}+\frac{5454}{4242}\)
\(=\frac{36}{35}+\frac{54}{42}\)
= \(\frac{36}{35}+\frac{9}{7}\)
= \(\frac{36}{35}+\frac{45}{35}\)
= \(\frac{81}{35}\)
a, 135 / 120 < 13/8
b, 3535/4848 > 5/8
c, 650650/480480 < 222222/144144
Tk cho mk nha!!!!
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
rút gọn: 2525/3535=5/7
505050/707070=5/7
=> =
rút gọn: 200520052005/200720072007= 2005/2007
=> >
a)\(\frac{2525}{3535}=\frac{5}{7}\)
\(\frac{505050}{707070}=\frac{5}{7}\)
=> \(\frac{2525}{3535}=\frac{505050}{707070}\)
b) \(\frac{200520052005}{200720072007}=\frac{2005}{2007}\)
vì \(\frac{2005}{2006}>\frac{2005}{2007}\)
nên \(\frac{2005}{2006}>\frac{200520052005}{200720072007}\)
C= 101 . 35 . 10101.23 / 35.10101 . 23. 101 = 1
Do D = 3535/3534 > 1 => C<D