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a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-1\right)\)
\(=\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.1=\dfrac{5}{9}\)
a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-\dfrac{26}{26}\right)\)
=\(\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.\left(\dfrac{7}{7}\right)\)
=\(\dfrac{5}{9}.1\)
\(=\dfrac{5}{9}\)

2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)

A=4/1.31+6/7.41+9/9.41+ 7/10.57
=20/35.31+30/35.41+45/45.41+35/50.57
=5(4/35.31+6/35.41+9/45.41+7/50.57)
=5(1/31-1/35+1/35-1/41+1/41-1/45+1/45-1/50+1/50-1/57)
=5(1/31-1/57)
B thì làm tương tự nhưng nhân với 2=> B=2(1/31-1/57)
=> A/B=5/2

\(A=\dfrac{-3}{5}.\dfrac{3}{13}+\dfrac{-3}{5}.\dfrac{10}{13}+5\)
\(A=\dfrac{-3}{5}\left(\dfrac{3}{13}+\dfrac{10}{13}\right)+5\)
\(A=\dfrac{-3}{5}.1+5\)
\(A=\dfrac{-3}{5}+5\)
\(A=\dfrac{22}{5}\)

b/\(\frac{10^9+1}{10^{9+1}+1}\)=\(\frac{10^9+1}{10.10^9+1}\)=\(\frac{1}{10\text{}}\)
\(\frac{10^{10}+1}{10^{10+1}+1}\)=\(\frac{10^{10}+1}{10+10^{10}+1}\)=\(\frac{1}{10}\)
Vì \(\frac{1}{10}\)=\(\frac{1}{10}\)=>bằng nhau

a) A = 23! + 19! + 15!
Vì số hạng trong A đều có thừa số là 10 nên các số ấy đều chia hết cho 10. \(\Rightarrow\) A \(⋮\) 10
\(\Rightarrow\) ĐPCM
b) Vì số hạng trong A đều có thừa số là 10 và 11 nên các số ấy đều chia hết cho 10 . 11 hay 110.
\(\Rightarrow\) A \(⋮\) 110
\(\Rightarrow\) ĐPCM

\(=\frac{23^{10}\left(1-23\right)}{23^{10}\left(23+21\right)}=\frac{-22}{44}=\frac{-1}{2}\)
\(\frac{23^{10}-23^{11}}{23^{11}+21.23^{10}}\)
\(=\frac{23^{10}.\left(1-23\right)}{23^{10}.\left(23+21\right)}\)
\(=\frac{-22}{44}=\frac{-22}{22.2}=\frac{-1}{2}\)
bằng 1e+23
Bằng 100000000000000000000000 đó bạn