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Ta thấy:
\(121\vdots11\\110\vdots11\\99\vdots11\\88\vdots11\\...\\11\vdots11\\\Rightarrow 121-110+99-88+...+11\vdots11\)
Để \(B=121-110+99-88+...+11+a\)\(⋮̸11\)
thì \(a⋮̸11\)
Mặt khác: a là số lẻ nhỏ hơn 10
\(\Rightarrow a\in\left\{1;3;5;7;9\right\}\)
a) 48.19 + 48.115 + 134.52
= 48.(19+115)+134.52
= 48.134+134.52
= 134.(48+52)
= 134.100
= 13400
$b.27.121+87.27+73.3= 27.(121+87)+73.34b) 27.121 + 87.27 + 73.34
= 27.(121+87)+73.34
= 27.208 + 73.34
= 5616 + 2482
= 8098
\(-\left(-239\right)+115+\left(-27\right)+\left(-125\right)-121\)
\(=239+115-27-125-121\)
\(=81\)
-(-239) + 115 + (-27) + (-125) - 121
= 239 + (115 - 125) + (-121 -27)
= 239 + (-10) + (-148)
= 91
Đáp án cần chọn là: B
x − 1 15 = 1 10 x = 1 10 + 1 15 x = 3 30 + 2 30 x = 5 30 x = 1 6
a ) 100000 : 10 3 = 10 5 : 10 3 = 10 2 b ) 11 5 : 121 = 11 5 : 11 2 = 11 3 c ) 243 : 3 3 : 3 = 3 5 : 3 3 : 3 = 3 1 d ) 4 8 : 64 : 16 = 4 8 : 4 3 : 4 = 4 4
A = \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + ... + \(\dfrac{1}{120}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + ... + \(\dfrac{1}{120}\))
A = \(2\).(\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\)... + \(\dfrac{1}{240}\))
A = 2.(\(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + ... + \(\dfrac{1}{15.16}\))
A = 2.(\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{15}\) - \(\dfrac{1}{16}\))
A = 2.(\(\dfrac{1}{4}\) - \(\dfrac{1}{16}\))
A = 2.\(\dfrac{3}{16}\)
A = \(\dfrac{3}{8}\)