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a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
1 a) (x+ 1) + (x + 2 ) + (x + 3) + ... + (x + 100) = 205550 (100 cặp)
=> (x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 205 550
100 số hạng x 100 số hạng
=> 100.x + 100 . 101 : 2 = 205 550
=> 100.x + 5050 = 205 550
=> 100 . x = 205 550 - 5050
=> 100 . x = 200500
=> x = 200500 : 100
=> x = 2005
a, 2006 x 2004 - \(\frac{2}{1995}\) + 2004 x 2005 = 8038043,999
b, 2006 x 125 + \(\frac{1000}{126}\) x 2006 - 1006 = 265664,6349
c, A = 1991 x 1999
=> A = ( 1995 - 4 ) x ( 1995 + 4 )
A = 1995 x ( 1995 + 4 ) - 4 x ( 1995 + 4 )
A = 1995 x 1995 + 1995 x 4 - ( 4 x 1995 + 4 x 4 )
A = 1995 x 1995 - 4 x 4
mà B = 1995 x 1995
Vậy A < B
d, Gọi giá trị biểu thức là C
C = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
C x 2 = \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)
C x 2 = \(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)
Vậy C x 2 - C = \(\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)\)
C = \(\frac{2}{3}-\frac{1}{96}\) ( vì phân số nào có ở số bị trừ cũng có ở số trừ thì trừ hết rồi nên không còn )
C = \(\frac{21}{32}\)
A=1991x1999=(1995-4)1999=1995x1999-4x1999
B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A
vậy A<B
A=1991x1999=
(1995-4)1999
=1995x1999-4x1999
B=1995x1995
=1995x(1999-4)
=1995x1999-1995x4>1995x1999-4x1999=A
vậy A<B
a, \(1-6+11-16+21-26+...+91-96+101\)\
\(\left(1+11+21+...+91+101\right)^{\left(1\right)}-\left(6+16+26+...+96\right)^{\left(2\right)}\)
Ta gọi (1) là B
(2) là A
Tổng dãy B là: ( 91 - 1) : 10 + 1 : 2 . ( 91 +1 ) + 101 = 561
Tổng dãy A là: ( 96 - 6) : 10 + 1 : 2 . ( 96 + 6 ) = 510
1 - 6 + 11 - 16 + 21 - 26 + ......... + 91 - 96 + 101 = 561 - 510
= 51
b, A = 1991 . 1999 = 1991 . ( 1995 + 4 ) = 1991 . 1995 + 1991 . 4
B = 1995 . 1995 = 1995 . ( 1991 + 4 ) = 1995 . 1991 + 1995 . 4
1991 < 1995 => A < B
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
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
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)
\(\frac{1999x1999}{1995x1995_{ }}=\frac{1999^2}{1995^2}=\left(\frac{1999}{1995}\right)^2\)\(>1^2\)\(=1\)
Bài 2 bấm máy tính nhé !