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#)Giải :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3A=3-1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(3A-A=\left(3-1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(3A-A=3-1-1+\frac{1}{243}=\frac{244}{243}\)
\(A=\frac{244}{243}:\left(3-1\right)=\frac{122}{243}\)
\(A=\frac{122}{243}\)
#~Will~be~Pens~#
Tìm y :
( y + 1/3 ) + ( y + 1/9 ) + ( y + 1/27 ) + ( y + 1/81 ) = 56/81
y + 1/3 + y + 1/9 + y + 1/27 + y + 1/81 = 56/81
y x 4 + ( 1/3 + 1/9 + 1/27 +1/81 ) = 56/81
y x 4 + ( 27/81 + 9/81 + 3/81 + 1/81 ) = 56/81
y x 4 + 40/81 = 56/81
y x 4 = 56/81 - 40/81
y x 4 = 16/81
y = 16/81 : 4
y = 4/81
a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50
=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3
3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)
3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60
3A = -2.3.4 + 49.60.61
\(A=\frac{-2.3.4+49.60.61}{3}=59772\)
b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53
=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6
6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)
6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53
6B = 1.3 + 51.53.55
\(B=\frac{1.3+51.53.55}{6}=24778\)
cau c mk ko bk
d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
D = 30+31+32+33+34+35+36+37+38
=> 3D = 31+32+33+...+38+39
=> 3D - D = 39-30
2D = 39-1
\(D=\frac{3^9-1}{2}=9841\)
C=1x2+2x3+3x4 +......+99x100
<=>3xC=1x2x3+2x3x3+.....+99x100x3
<=>3xC=1x2x3+2x3x(4-1)+.....+99x100x(101-98)
<=>3xC=1x2x3-1x2x3+2x3x4-2x3x4+......-98x99x100+99x100x101
<=>3xC=99x100x101 <=> C=99x100x101 :3 <=> C= 333300
\(C=1.2+2.3+3.4+...+99.100\)
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 99.100.3
3C = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ...+ 99.100.(101-98)
3C = 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + ...+ 99.100.101-98.99.100
\(3C=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3C=99.100.101\Rightarrow C=\frac{99.100.101}{3}=333300\)
Dấu . là dấu nhân
3C = 3 x C
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
bạn quy đồng mẫu số tất cả cả các phân số trên lên thành 243 rồi cộng vào là được
\(C=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(C=\frac{81}{243}+\frac{27}{243}+\frac{9}{243}+\frac{3}{243}+\frac{1}{243}\)
\(C=\frac{81+27+9+3+1}{243}=\frac{121}{243}\)
Vậy C = ...