=20/27

bằng \(\frac{20}{27}\)

1212/1515 :2727/2525=4/5:27/25=4/5.25/27=20/27

Học tốt!

\(\frac{1212}{1515}:\frac{2727}{2525}\)

\(=\frac{1212}{1515}x\frac{2525}{2727}\)

\(=\frac{20}{27}\)

Chúc Học tốt

\(\frac{1212}{1515}:\frac{2727}{2525}=\frac{1212}{1515}x\frac{2525}{2727}=\frac{12x101}{15x101}x\frac{25x101}{27x101}=\frac{12x25}{15x27}=\frac{3x4x5x5}{3x5x27}=\frac{4x5}{27}=\frac{20}{27}.\)

Thanks

bước 1: rút về số tối giản

bước 2: tìm MSC

bước 3: tính...

\(\frac{1212}{1515}:\frac{2727}{2525}\)

Rút gọn : \(\frac{1212}{1515}=\frac{4}{5}\)

\(\frac{2727}{2525}=\frac{27}{25}\)

\(=>\frac{4}{25}.\frac{25}{27}=\frac{20}{27}\)

~ chúc bn hok tốt ~

cám ơn nha

a) \(\frac{1212}{1515}\):\(\frac{2727}{2525}\)

=   \(\frac{1212}{1515}\)* \(\frac{2525}{2727}\)

=     \(\frac{101.12}{101.15}\)* \(\frac{101.25}{101.27}\)

=      \(\frac{12}{15}\). \(\frac{25}{27}\)

=       \(\frac{20}{27}\)

b) ban co the viet ro hon de bai dc ko?

a)=4.25x(58.47+41.53)-125

=4.25x100-125

=425-100

=325

=15/23.(69/54./27/15)

=15/23.23/10

=3/2

\(=\dfrac{15}{23}\times\dfrac{69}{54}\times\dfrac{27}{15}\\ =\dfrac{3}{2}\)

=15/23 x (69/54 x /27/15)

=15/23 x 23/10

=3/2

2234234234 ákdsdaslhfsodf

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)

=\(\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)

=\(\dfrac{16}{11}\)

Giải:

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\) 

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\) 

\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\) 

\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\) 

\(=6.\dfrac{8}{33}\) 

\(=\dfrac{16}{11}\)

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}=12\cdot\left[\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right]=12\cdot\left[\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}\right]=12\cdot\dfrac{1}{2}\cdot\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right]=6\cdot\left[\dfrac{1}{3}-\dfrac{1}{11}\right]=6\cdot\dfrac{8}{33}=\dfrac{48}{33}=\dfrac{16}{11}\)

12/27+a=25/50

4/9+a=5/10

a=5/10-4/9

a=5/90