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a)82*324=(23)2*(25)4=26*220=226
b)273*94*243=(33)3*(32)4*(3*8)3=39*38*33*83=320*83
c)77+6*49*75=75+6*72*75=75+6*77=75(1+6*72)
Bài 2 : Tính
1/ (-15)-/-10/+/-9/-/-5/
=-15-10+9-5
=-21
2/14-(-13)-(-21)+(-32)
=14+13+21-32
=16
3/(-12)-(-17)-(-21)+(-32)
= -12+17+21-32
=-6
4/-/-14/ + /-10/-(-12)+(-8)
=-14+10+12-8
=0
5/-(-11)+(-15)+(+13)-21
=11-15+13-21
=-12
CÂU16
a,5[-8]2[-3] b,3[-5]2+2[-5]-20 c,34[15-10]-15[34-10]
=5.2{-8.[-3]} =3. 25+{[-10]-20} =34.15-340-15. 34-150
=10. 24 =75+[-30] =0+[-340-150]
=2400 =45 =-490
d,27[-17]+[-17]73 e, 512[2-128]-128[-512]
=-17[27+73] =512. [-126]-128[-512]
=-17. 100 =-64512- [-65536]
-1700 =1024
CÂU 17
a,5-[10-x]=7 b, [4x-2][x+5]=0 c,2x-9=-8-9
5-10+x=7 ⇒4x-2 hoặc x+5=0 2x-9=-17
x=7-5+10 TH1:4x-2=0 TH2 x+5=0 2x=-17+9
x=12 4x=0+2 x=0-5 2x=-8
4x=2 x=-5 x=-8:2
x=2:4 x=-4
x=2/4 ko thỏa mãn vì x∈Z
Vậy x=-5
d,3[x-1]-27=0 e,5[3x+8]-7[2x+3]=16
3[x-1]=0+27 15x+40-14x+21=16
3[x-1]=27 15x-14x=16-21-40
[x-1]=27-3 x=-15
[x-1]=24
⇒x-1 =24 hoặc -24
TH1:x-1 =24 TH2 :x-1 =-24
x=24+1 x=-24+1
x=25 x=-23
Vậy x=25 hoặc -23
Bài 1:
a ) \(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
\(x+\frac{1}{9}=\frac{3}{6}+\frac{3}{5}\)
\(x+\frac{1}{9}=\frac{11}{10}\)
\(x=\frac{11}{10}-\frac{1}{9}=\frac{89}{90}\)
Vậy \(x=\frac{89}{90}\)
b) \(\frac{3}{4}-x+\frac{6}{11}=\frac{5}{6}\)
\(\frac{3}{4}-x=\frac{5}{6}-\frac{6}{11}\)
\(\frac{3}{4}-x=\frac{19}{66}\)
\(x=\frac{3}{4}-\frac{19}{66}=\frac{61}{132}\)
Vậy \(x=\frac{61}{132}\)
Bài 2 :
a) \(x:\frac{13}{16}=\frac{5}{-8}\)
\(x=\frac{5}{-8}.\frac{13}{16}=-\frac{65}{128}\)
Vậy \(x=-\frac{65}{128}\)
b) \(x.\frac{-14}{28}=\frac{6}{-9}-\frac{2}{15}\)
\(x.\frac{-14}{28}=-\frac{4}{5}\)
\(x=-\frac{4}{5}:\frac{-14}{28}=\frac{8}{5}\)
Vậy \(x=\frac{8}{5}\)
\(1)C=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\)
\(3C=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\)
\(3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\right)\)
\(2C=1-\dfrac{1}{162}\)
\(2C=\dfrac{161}{162}\)
\(C=\dfrac{161}{162}.\dfrac{1}{2}\)
\(C=\dfrac{161}{324}\)
\(2)A=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\right)\)
\(A=1-\dfrac{1}{512}=\dfrac{511}{512}\)