1.Tính tổng sau:
\(\frac{1919}{3535}\)+ \(\frac{54545454}{75757575}\)
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LT
3
20 tháng 6 2018
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 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
VN
15 tháng 5 2019
Mình giải theo kiểu lớp 6 nhá !
=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)
=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)
=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)
=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)
=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)
=\(33.\frac{136}{33}\)
=\(\frac{33.136}{33}\)(*)
=\(136\)
(Bạn có thể bỏ bước có dấu *)
HL
1
4 tháng 5 2019
( 3434/3535 + 3434/6363 + 3434/9999 ) × 55
= ( 34/35 + 34/63 + 34/99 ) × 55
= 34 × ( 1/35 + 1/63 + 1/99 ) × 55
= 34 × 3/55 × 55
= 102/55 × 55
= 102
Mình ko chắc lắm đâu nhá
LM
12 tháng 7 2017
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,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
NA
1
21 tháng 6 2017
C= 101 . 35 . 10101.23 / 35.10101 . 23. 101 = 1
Do D = 3535/3534 > 1 => C<D
8 tháng 8 2019
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\cdot\frac{4}{33}\)
\(=\frac{16}{11}\)
L
8 tháng 8 2019
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\left(\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\)
\(=12\cdot\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{3\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=6\cdot\frac{8}{33}\)
\(=\frac{48}{33}\)
PT
1
HL
3
30 tháng 4 2018
1616/1515 + 1616/3535 + 1616/6363 + 1616/9999
= 32/21 + 1616/6363 + 1616/9999
= 16/9 + 1616/9999
= 64/33
\(\frac{221}{175}\)
1919/3535+54545454/75757575
=19/35+54/75=221/175