Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
Câu a:
A = \(\frac{1}{2\times3}\) + \(\frac{1}{3\times4}\) + \(\frac{1}{4\times5}\) + \(\frac{1}{5\times6}\) + \(\frac{1}{6\times7}\) + \(\frac{1}{7\times8}\)
A = \(\frac12-\frac13\) + \(\frac13-\frac14\) + \(\frac14-\frac15\) + \(\frac15-\frac16\) + \(\frac16-\frac17\) + \(\frac17-\frac18\)
A = \(\frac12-\frac18\)
A = \(\frac38\)
Câu b:
A = \(\frac12\) + \(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\) + \(\frac{1}{256}\)
2 x A = 1 + \(\frac12\) + \(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\)
2 x A - A = 1 + \(\frac12\) +\(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\) - \(\frac12-\frac14\) -...-\(\frac{1}{128}\) -\(\frac{1}{256}\)
A x (2 - 1) = (1 - \(\frac{1}{256}\)) + (\(\frac12\)-\(\frac12\)) +...+(\(\frac{1}{128}\) - \(\frac{1}{128}\))
A = 1 - \(\frac{1}{256}\) + 0 + 0+...+ 0
A = \(\frac{255}{256}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128
A=1-1/128
A=127/128
Vậy A=\(\frac{127}{128}\)
B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
B=1-1/100
B=99/100
Vậy B=\(\frac{99}{100}\)
Ta có :
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
\(C=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(1-\frac{1}{14}\right)\)
\(C=1-\frac{1}{14}\)
\(C=\frac{14}{14}-\frac{1}{14}\)
\(C=\frac{14-1}{14}\)
\(C=\frac{13}{14}\)
Vậy \(C=\frac{13}{14}\)
Chúc bạn học tốt ~
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{13\cdot14}\)
\(C=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+....+\frac{14-13}{13\cdot14}\)
\(C=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+....+\frac{14}{13\cdot14}-\frac{13}{13\cdot14}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{13}-\frac{1}{14}\)
\(C=1-\frac{1}{14}\)
\(C=\frac{13}{14}\)
dấu "." là dấu nhân nhs
Mấy câu khác bạn tự tính nhé, dễ thôi. Câu 2, câu 3 giống dạng nhau. Dấu . có nghĩa là dấu nhân nhé.
Câu 2:
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
Câu 4 b) bạn quy đồng 2 phân số cho cùng mẫu rồi tìm thôi.
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{4}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{12}{9}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
Từ chỗ này làm dễ hơn rồi bạn tự làm tiếp đi nhé
a) đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2\times A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2\times A-A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
phần b bn cx lm tương tự như z nha!
c) sửa đề:
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{13x14}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
\(=1-\frac{1}{14}=\frac{13}{14}\)
Sửa:
d) \(\frac{1}{15x18}+\frac{1}{18x21}+\frac{1}{21x24}+...+\frac{1}{87x90}\)
\(=\frac{1}{3}x\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(=\frac{1}{3}x\frac{1}{18}\)
\(=\frac{1}{54}\)