tính nhanh
a,321/963+234/468 234/468-666/1998=
b,1/2+1/4+1/8+1/16+1/32+1/64+1/128=
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Có tên giống mk quá!
1 Tính tổng
s = 9 + 99 + 999...+.....+9( 100 số 9)
s = 9 + 9 x 11+ 9 x 111 + 9 x 111 x11(100 chữ số 1)
s = 9 x ( 1+ 11 +111+1111+......+1111..11)(100 chữ số 1)
s = 9 x 111.....1111
s = 999...99
Bài 2 mk k bít làm thông cảm nha.Nhưng mk gửi lời mời kb.
1,
S=(10-1)+(10^2-1)+(10^3-1)+....+(10^100-1).
=(10+10^2+10^3+...+10^100)-10^2(chữ số 1)
10*S=10^2+10^3+10^4+...+10^101-10^3.
=>10*S-S=(10^2+10^3+...+10^101-10^3)-(10+10^2+10^3+...+10^100-10^2).
=>9S=10^101-10^3-10+10^2
9S=10^101-910
S=(10^101-910)/9.
2,
Tử:Mẫu=(2839+2162-1)*468/468=5000 (1)
Và:
1-1/2=1/2
1-1/3=2/3
1-1/4=3/4
1-1/5=4/5
.........
1-1/100=99/100
Nhân lại với nhau,rút gọn ta được: 1/100 (2)
Từ (1),(2)=> a=5000*1/100=50.
a,234.4+234.5+234=234(4+5+1)=234.10=2340
b,135.16-135.2-135.4=135(16-4-2)=135.10=1350
c,1/2+1/4+1/8+1/16+1/32+1/64=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32=1-1/32=31/32
a) 234 x4 + 234 x 5 + 234 = 234 x ( 1+4+5)=2340
b)135 x16 -135 x2 -4 x135 =135 x (16-2-4) =1350
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)
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=\(\frac{99}{100}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A\cdot2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\) \(-\) \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=\) \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
Ta lấy\(\frac{1}{128}\)là MSC. Ta tính được \(\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
Kết quả bằng \(\frac{127}{128}\)
1/2 + 1/4 + 1/8 + 1/16 +1/32 + 1/64 + 1/128
=1-1/2+1/2-1/4+1/4-1/8+...+1/64+1/128
=1-1/128
=127/128
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
A=\(\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)=\frac{1}{2}\left(1+A-\frac{1}{2}-\frac{1}{128}\right)\)
2A=\(A+\frac{1}{2}-\frac{1}{128}=A+\frac{63}{128}\)
=> A=\(\frac{63}{128}\)
A = \(\dfrac{321}{963}\) + \(\dfrac{234}{468}\) - \(\dfrac{666}{1998}\)
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
A = 1/2
B = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2xB = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2x B - B = 1 - \(\dfrac{1}{128}\)
B = \(\dfrac{127}{128}\)