Biết rằng số có 6 chữ số A9876A chia hết cho 9 và 2 tìm giá trị của A.
A. 1 B. 2 C. 3 D. 4
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Tử số:
\(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{100}\)
\(=(\dfrac{1}{2}+\dfrac{1}{100})+(\dfrac{1}{4}+\dfrac{1}{98})+...+\left(\dfrac{1}{48}+\dfrac{1}{54}\right)+(\dfrac{1}{50}+\dfrac{1}{52})\)
\(=\dfrac{102}{2.100}+\dfrac{102}{4.98}+...+\dfrac{102}{48.54}+\dfrac{102}{50.52}\)
\(=102\left(\dfrac{1}{2.100}+\dfrac{1}{4.98}+...+\dfrac{1}{48.54}+\dfrac{1}{50.52}\right)\)
Mẫu số:
\(\dfrac{1}{2.100}+\dfrac{1}{4.98}+...+\dfrac{1}{98.4}+\dfrac{1}{100.2}\)
\(=\dfrac{2}{2.100}+\dfrac{2}{4.98}+....+\dfrac{2}{48.54}+\dfrac{2}{50.52}\)
\(=2\left(\dfrac{1}{2.100}+\dfrac{1}{4.98}+...+\dfrac{1}{48.54}+\dfrac{1}{50.52}\right)\)
Chia cả tử và mẫu cho \(=\dfrac{1}{2.100}+\dfrac{1}{4.98}+...+\dfrac{1}{48.54}+\dfrac{1}{50.52}\)
ta được \(\dfrac{tử}{mẫu}=\dfrac{102}{2}=51\)
Đs:....
b, \(\dfrac{x}{3}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{5}\)
\(\dfrac{x}{3}\) = \(\dfrac{1}{5}\) + \(\dfrac{1}{2}\)
\(\dfrac{x}{3}\) = \(\dfrac{7}{10}\)
x = \(\dfrac{7}{10}\) x 3
x = 21/10
c, x + \(\dfrac{1}{2}\) = \(\dfrac{6}{10}\)
x = \(\dfrac{6}{10}\) - \(\dfrac{1}{2}\)
x = \(\dfrac{1}{10}\)
d, x + \(\dfrac{3}{15}\) = \(\dfrac{1}{3}\)
x = \(\dfrac{1}{3}\) - \(\dfrac{3}{15}\)
x = \(\dfrac{2}{15}\)
h, \(\dfrac{4}{5}\) + x = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) - \(\dfrac{4}{5}\)
x = \(\dfrac{-2}{15}\)
\(\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}-\dfrac{4}{3}\)
\(=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)-\dfrac{4}{3}\)
\(=\dfrac{1}{3}.2-\dfrac{4}{3}\)
\(=\dfrac{2}{3}-\dfrac{4}{3}=-\dfrac{2}{3}\)
1/3 . 4/5 + 1/3 . 6/5 - 4/3
= 1/3 . ( 4/5 + 6/5 ) - 4/3
= 1/3 . 10 - 4/3
=10/3 - 4/3
=6/3
a) \(A=\dfrac{x+5}{x+1}=1+\dfrac{4}{x+1}\)
Để \(A\in Z\Rightarrow\left(x+1\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3;3;-5\right\}\)
b) \(B=\dfrac{2x+4}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=2-\dfrac{2}{x+3}\)
Để \(B\in Z\Rightarrow\left(x+3\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)
c) \(C=\dfrac{3x+8}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=3+\dfrac{11}{x-1}\)
Để \(C\in Z\Rightarrow\left(x-1\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ =\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)-...-\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{2}\right)\\ =\dfrac{1}{2014}+\dfrac{1}{2014}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2012}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\\ =\dfrac{1}{2014}+\dfrac{1}{2014}-1\\ =\dfrac{1}{1007}-1\\ =-\dfrac{1006}{1007}\)
Lấy hiệu hai phân số rồi suy ra điều phải chứng minh
\(\dfrac{-214}{317}-\dfrac{-21}{38}=\dfrac{21.317-214.38}{317.38}=\dfrac{-1475}{317.38}< 0\)
\(\Rightarrow\dfrac{-214}{317}< \dfrac{-21}{38}\)
\(\dfrac{23}{45}>\dfrac{23}{46}=\dfrac{1}{2}\)
\(\dfrac{75}{151}< \dfrac{75}{150}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{23}{45}>\dfrac{75}{151}\)
Gọi a là chiều rộng đáy của hình hộp chữ nhật.
Ta có:
\(2.60.a=2.12.a+2.12.60\Leftrightarrow120a=24a+1440\)
\(\Leftrightarrow96a=1440\Rightarrow a=15\)
Vậy chiều rộng đáy của hình hộp chữ nhật là 15 cm
\(\overline{A9876A}⋮2\) => A chẵn
\(\overline{A9876A}⋮9\Rightarrow2A+9+8+7+6=2A+30=2A+3+27⋮9\)
\(27⋮9\Rightarrow2A+3⋮9\)
A chẵn => 2A+3 lẻ
\(A\le9\Rightarrow2A+3\le21\)
\(\Rightarrow2A+3=9\Rightarrow A=3\)
A9876A nha. Mình viết nhầm