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a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\)
\(=\left(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}\right)+\left(\dfrac{14}{35}+\dfrac{25}{35}-\dfrac{4}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+2+1}{6}+\dfrac{14+25-4}{35}+\dfrac{1}{41}\)
\(=1+1+\dfrac{1}{41}\)
\(=2+\dfrac{1}{41}\)
\(=2\dfrac{1}{41}=\dfrac{83}{41}\)
C=0.5+1/3+0.4+5/7+1/6-4/35+1/41
C=1/2+1/3+2/5+5/7+1/6-4/35+1/41
C=(1/2+1/6+1/3)+(2/5+5/7-4/35)+1/41
C=1+1-1/41
C=2-1/41
=>C=81/41
D=1/90-1/72-1/56-1/42-1/30-1/20--1/12-1/6-1/2
=> D=1/9*10-1/8*9-1/7*8-1/6*7-1/5*6-1/4*5-1/3*4-1/2*3-1/1*2
=>D=-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=>D=-(1-1/10)
=>D=-9/10
ai k mh mh k lại
a) Ta có: \(D=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{96}\)
\(=\dfrac{63}{96}=\dfrac{21}{32}\)
b)
Sửa đề: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
Ta có: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{2048}-\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{E}{2}=\dfrac{1}{2}-\dfrac{1}{4096}=\dfrac{2047}{4096}\)
hay \(E=\dfrac{2047}{2048}\)
Ta có
C=\(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(\left(=\right)C=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{2}{5}+\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(\left(=\right)C=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)+\left(\frac{14}{35}+\frac{25}{35}-\frac{4}{35}\right)+\frac{1}{41}\)
\(\left(=\right)c=1+\left(-1\right)+\frac{1}{41}=\frac{1}{41}\)
vì viết dưới dạng phân số chưa quen nên bn thông cảm cho mk nhé!
C = 0,5 + 1/3 + 0,4 + 5/7 + 1/6 - 4/35 + 1/41
= 5/10 + 1/3 + 4/10 + 5/7 + 1/6 - 4/35 + 1/41
= 1/2 + 1/3 + 2/5 + 5/7 + 1/6 - 4/35 + 1/41
= (1/2 + 1/3 + 1/6) + (2/5 + 5/7 - 4/35) +1/41
= (3/6+ 2/6 + 1/6) + ( 14/35 + 25/35 - 4/35) + 1/41
= 6/6 + 35/35 + 1/41
= 1 + 1 + 1/41 = 2 + 1/41 = 2/1 + 1/41
= 43/41 + 1/41 = 44/41
chúc bn học tốt
1 .
[ 6 + ( 1/2 ) 3 - /-0,5/ ] : 3/12
= [ 6 + 1/8 - ( -5/10 ) ] . 12/3
= [ 6 + 1/8 + 5/10 ] .4
= [ 6 + 1/8 + 1/2 ] .4
= [ 48/8 + 1/8 + 4/8 ] .4
= 53/8 . 4
= 53 . 1/2
= 53/2
2.
1/3 + 5/4 : (-35/16)
= 1/3 + 5/4 . ( 16/-35 )
= 1/3 + 1 . 4/4 . (-7 )
= 1/3 + 4/28
= 1/3 + 1/7
= 7/21 + 3/21
= 10/21
3 .
11 3/13 - ( 2 4/7 + 5 3/13 )
= 11 3/13 - 2 4/7 - 5 3/13
= { [ ( 11 - 5 ) . ( 3/13 - 3/13 ) } - 2 4/7
= ( 6 . 0 ) - 2 4/7
= 0 - 18/7
= 18/7
4 .
1 3/7 + (-1/3 + 2 4/7 )
= 1 3/7 + ( -1 )/3 + 2 4/7
={ [ ( 1 + 2 ) . ( 3/7 + 4/7 ) } + ( -1 )/3
= ( 3 . 1 ) + ( -1 )/3
= 3 + ( -1 )/3
= 9/3 + ( -1 )/3
= 8/3
5 .
( 6 4/9 + 3 7/11 ) - 4 4/9
= 6 4/9 + 3 7/11 - 4 4/9
= { ( 6 - 4 ) . ( 4/9 - 4/9 ) } + 3 7/11
= ( 2 . 0 ) + 3 7/11
= 0 + 40/11
= 40/11
Nếu đúng thì k cho mình nha !
\(C=0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}+\frac{1}{41}\)
\(=\frac{7}{5}+\frac{3}{5}+\frac{1}{41}\)
\(=2+\frac{1}{41}=\frac{83}{41}\)
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(\frac{1}{90}-\frac{1}{30}-\frac{1}{6}-\frac{1}{2}\right)+\left(-\frac{1}{72}-\frac{1}{12}\right)-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{1-2-15-45}{90}+\frac{-1-6}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=-\frac{61}{90}-\frac{7}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{-1708-245-45-60}{2520}\)
\(=-\frac{49}{60}\)
Nghịch đảo của C là \(\frac{41}{83}\), nghịch đảo của D là \(-\frac{60}{49}\)
\(\frac{41}{83}\cdot\left(-\frac{60}{49}\right)=-\frac{2460}{4067}\)
0,5+31+0,4+75+61−354+411
=12+13+25+57+16−435+141=21+31+52+75+61−354+411
=(12+13+16)+(25+57−435)+141=(21+31+61)+(52+75−354)+411
=(36+26+16)+(1435+2535−435)+141=(63+62+61)+(3514+3525−354)+411
=3+2+16+14+25−435+141=63+2+1+3514+25−4+411
=1+1+141=1+1+411
=2+141=2+411
=2141=8341=2411=4183