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1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243
tick đúng nha
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(3B-B=1-\frac{1}{243}\)
\(2B=\frac{242}{243}\)
\(B=\frac{242}{243}\div2\)
\(B=\frac{121}{243}\)
a.A=1/2+1/4+1/8+1/16+1/32+1/64
A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}\)
= 1 - 1/8 = 7/8
b.B=1/3+1/9+1/27+1/81+1/243
B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)
= 1 - 1/27 = 26/27
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2\times A=1-\frac{1}{729}=\frac{728}{729}\)
\(A=\frac{364}{729}\)
0,63 x 8 + 0,63 + 0,63 + 9,89
=0,63 x 8 + 0,63 x 1 + 0,63 x1 + 9,89
= 0,63 x ( 8+ 1 + 1 ) +9,89
= 0,63 x 10 +9,89
= 6,3 +9,89
= 16,19
a, 23/4 : 3 + 9/4 x 1/3 - 3/8
= 7,8 + 12,22 - 3,8
= 20,02 - 3,8
=16,22
b, 3/5 : 5/6 : 6/7 : 7/8 + 2/5 +23/35
=3/5 x 6/5 x 7/6 x 8/7 + 2/5 + 23/35
=24/25 + 2/5 + 23/35
=1/5 x(24/5 + 2 +23/7)
=1/5 x 353/35
=353/175
A = \(\dfrac{36\times950+18\times726\times2+3\times324\times12}{1+3+5+7+...+27+29+31-152}\)
A = \(\dfrac{36\times950+36\times726+36\times324}{1+3+5+7+...+27+29+31-152}\)
MS = 1 + 3 + 5 +...+ 27 + 29 + 31 - 152
Xét dãy số: 1; 3; 5;...;27; 29; 31 là dãy số cách đều với khoảng cách là:
3 - 1 = 2
Số số hạng của dãy số là:
(31 - 1): 2 + 1 = 16
MS = (31 + 1) x 16 : 2 - 152 = 104
A = \(\dfrac{36\times\left(950+726+324\right)}{104}\)
A = \(\dfrac{36\times2000}{104}\)
A = \(\dfrac{9000}{13}\)
\(a.=9.3\times\left(6.7+3.3\right)\)
\(=9.3\times10\)
\(=93\)
\(b.=0.35\times\left(7.8+2.2\right)\)
\(=0.35\times10\)
\(=3.5\)
(\(\dfrac{2}{3}\) + \(\dfrac{8}{9}\) + \(\dfrac{26}{27}\) + \(\dfrac{80}{81}\) + \(\dfrac{242}{243}\)) : y = 5
Đăt A = \(\dfrac{2}{3}\) + \(\dfrac{8}{9}\) + \(\dfrac{26}{27}\) + \(\dfrac{80}{81}\) + \(\dfrac{242}{243}\)
3A = 2 + \(\dfrac{8}{3}\) + \(\dfrac{26}{9}\) + \(\dfrac{80}{27}\) + \(\dfrac{242}{81}\)
3A - A = 2 + \(\dfrac{8}{3}\) + \(\dfrac{26}{9}\) + \(\dfrac{80}{27}\) + \(\dfrac{242}{81}\) - \(\dfrac{2}{3}\)-\(\dfrac{8}{9}\)-\(\dfrac{26}{27}\)-\(\dfrac{80}{81}\)-\(\dfrac{242}{243}\)
A x (3 - 1) = 2 - \(\dfrac{242}{243}\)+ (\(\dfrac{8}{3}\) - \(\dfrac{2}{3}\))+(\(\dfrac{26}{9}\) - \(\dfrac{8}{9}\))+(\(\dfrac{80}{27}\)-\(\dfrac{26}{27}\))+(\(\dfrac{242}{81}\)-\(\dfrac{80}{81}\))-\(\dfrac{242}{243}\)
A x 2 = 2 - \(\dfrac{242}{243}\) + 2 + 2 + 2 + 2
A x 2 = (2 + 2 + 2 +2 + 2) - \(\dfrac{242}{243}\)
A x 2 = 2x5 - \(\dfrac{242}{243}\)
A x 2 = 10 - \(\dfrac{242}{243}\)
A x 2 = \(\dfrac{2188}{243}\)
A = \(\dfrac{2188}{243}\) : 2
A = \(\dfrac{1094}{243}\)
\(\dfrac{1094}{243}\) : y = 5
y = \(\dfrac{1094}{243}\) : 5
y = \(\dfrac{1094}{1215}\)