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DT
Đỗ Thanh Hải
CTVVIP
17 tháng 3 2021
a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)
Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)
T
12 tháng 8 2023
a) \(\dfrac{3}{7}-\dfrac{4}{11}-\dfrac{3}{7}\)
=\(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)-\dfrac{4}{11}\)
\(=-\dfrac{4}{11}\)
b) \(-\dfrac{5}{11}+\dfrac{3}{8}-\dfrac{6}{11}\)
\(=\left(-\dfrac{5}{11}-\dfrac{6}{11}\right)+\dfrac{3}{8}\)
\(=-1+\dfrac{3}{8}=-\dfrac{8}{8}+\dfrac{3}{8}=-\dfrac{5}{8}\)
c) \(\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{11}{4}\)
\(=\left(\dfrac{3}{4}-\dfrac{11}{4}\right)-\dfrac{1}{3}\)
\(=-2-\dfrac{1}{3}=-\dfrac{6}{3}-\dfrac{1}{3}=-\dfrac{7}{3}\)
KV
12 tháng 8 2023
a) 3/7 - 4/11 - 3/7
= (3/7 - 3/7) - 4/11
= -4/11
b) -5/11 + 3/8 - 6/11
= (-5/11 -6/11) + 3/8
= -1 + 3/8
= -5/8
c) 3/4 - 1/3 - 11/4
= (3/4 - 11/4) - 1/3
= -2 - 1/3
= -7/3
12 tháng 2 2018
a ) \(\frac{-4}{11}\)x \(\frac{5}{15}\)x \(\frac{11}{-4}\)
= \(\frac{-4.5.11}{11.15.\left(-4\right)}\)
= \(\frac{1}{3}\)
TM
8 tháng 4 2020
a) 5/9 + 4/9 . 3/7 + 4/9 . 4/7
= 5/9 + 4/9 . (3/7 + 4/7)
= 5/9 + 4/9 . 1
= 5/9 + 4/9
= 1
18 tháng 3 2017
\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
27 tháng 2 2022
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
12 tháng 8 2018
a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
C = 4 + 4^2 + 4^3 + ... + 4^11
4C = 4^2 + 4^3 + 4^4 + ... + 4^12
=> 4C - C = 3C = 4^12 - 4
=> C = (4^12 - 4) : 3
C=\(4+4^2+4^3+...+\)\(4^{11}\)
=> 4C=\(4^2+4^3+4^4+...+4^{12}\)
=> 4C-C=\(4^2+4^3+4^4+...+4^{12}\)-\(4-4^2-4^3-...-4^{11}\)
=> 3C=\(4^{12}-4\)=> C=\(\frac{4^{12}-4}{3}\)
vậy C=\(\frac{4^{12}-4}{3}\)