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Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
2046-(47.48-47.47-20-27)
=2046-(47-20-27)
=2046
2+4+6+8+16+32+.....+512+1024
bài còn lại mình đang suy nghĩ nhé
bao giờ ra mình gửi cho
Đừng vì thế mà ko đánh k cho mình nha
a)
\(A=2+4+8+...+2048\)
\(A=2+2^2+...+2^{11}\)
\(2A=2^2+2^3+...+2^{12}\)
\(2A-A=\left(2^2+2^3+...+2^{12}\right)-\left(2+2^2+...+2^{11}\right)\)
\(A=2^{12}-2\)
a) 2+4+8+16+32+64+128+512+1024+2048
b thì minh cha ra. Mình sẽ cố làm ra b mong ban thong cam va bạn nho k đung cho minh nha
\(2+4+8+16+...+1024+2048=2\left(1+2+4\right)+16\left(1+2+4\right)+128\left(1+2+4\right)+1024\left(1+2\right)=\left(2+16+128+1024\right)\left(1+2+4\right)+3072=1022+3072=4094\)
A = 2+4+8+16+...+1024+2048
=> A = 2 + 22 + 23 + ... + 211
=> 2A = 22 + 23 + 24 ... + 212
=> 2A - A = 22 + 23 + 24 ... + 212 - 2 + 22 + 23 + ... + 211
=> A = 212 - 2
Số số hạng là:
( 2048 - 2 ) : 2 + 1 = 1024 ( số )
Tổng là:
( 2048 + 2 ) x 1024 : 2 = 1 049 600
\(A=2+4+8+...+1024+2048\)
\(A=2+2^2+2^3+...+2^{10}+2^{11}\)
\(2A=2^2+2^3+...+2^{12}\)
\(2A-A=2^2+2^3+2^4+...+2^{12}-\left(2+2^2+...+2^{11}\right)\)
\(A=2^{12}-2=4094\)
còn có 1 cách tiểu học nhưng cũng gần giống thế này:
\(A=2+4+8+...+1024+2048\)
\(2A+2=\frac{2+4+8+...+1024+2048}{A}+4096\)
\(\Rightarrow A+2=4096\Rightarrow A=4094\)
no way
LÊN GÕ MÁY TÍNH CÁCH GIẢI NHANH NHẤT BN MUỐN