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a) ab=a.10+b
ba=b.10+a
ab-ba=10a+b-10b-a
=9a-9.b
Giả sử a lớn hơn b n đơn vị, ta có:
(b+n)9-9b
=n.9 => ab-ba luôn chia hết cho 9
b) ab=10a+b
ba=10b+a
ab+ba=10a+a+10b+b
=11a+11b
=(a+b)11
=> ab+ba luôn chia hết cho 11
chúc bạn học tốt nha
Ta có: ab - ba = 10a + b - (10b + a) = 10a + b - 10b - a = 9a - 9b = 9 x (a - b)
Vì a > b nên a - b dương => 9 x (a - b) chia hết cho 9
ab + ba = 10a + b + 10b + a = 11a + 11b = 11 x (a + b) chia hết cho 11
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)
A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)
A \(\times\)( 2-1) = \(\dfrac{255}{128}\)
A = \(\dfrac{255}{128}\)
Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T
\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)
\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(T=2+0+0+...-\dfrac{1}{128}\)
\(T=\dfrac{256}{128}-\dfrac{1}{128}\)
\(T=\dfrac{255}{128}\)
\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{512}+\frac{1}{1024}=??????????\)
\(< =>1+\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+...+\frac{1}{2\cdot256}+\frac{1}{2\cdot512}\)
\(< =>1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{256}+\frac{1}{2}-\frac{1}{512}\)
\(< =>1+\frac{1}{1}-\frac{1}{512}\)
\(< =>\frac{1023}{512}\)
chuc ban hoc tot nhe :))
A = 3 + 32 ....+ 330
A = (3 + 32 + 33) + (34 + 35 + 36) +...+ (328 + 329 + 330)
A = 3.( 1 + 3 + 32) + 34.( 1 + 3 + 32) +...+ 328.(1 + 3 + 32)
A = (1+3+32).( 3 + 34 + ...+ 328)
A = 13.(3 +34 +...+ 328)
13 ⋮ 13 ⇒ A = 13.(3 + 34+...+328) ⋮ 13 (đpcm)
1/128 * 2 = 1/64
1/64 * 2 = 1/32
1/32 * 2 = 1/16
...
1/4 * 2 = 1/2
1/2 * 2 = 1
Mà A chỉ có một số hạng 1/128 nên tính ra được 127/128
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)