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A = 3/1×5 + 3/5×9 + 1/9×13 + ... + 9/97×101 + 3/101×105
A = 3/4 × (4/1×5 + 4/5×9 + 4/9×13 + ... + 4/97×101 + 4/101×105)
A = 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101 + 1/101 - 1/105)
A = 3/4 × (1 - 1/105)
A = 3/4 × 104/105
A = 26/35
B = 1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625
5B = 1 + 1/5 + 1/25 + 1/125 + 1/625 + 1/3125
5B - B = (1 + 1/5 + 1/25 + 1/125 + 1/625 + 1/3125) - (1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625)
4B = 1 - 1/15625
4B = 15624/15625
B = 15624/15625 : 4
B = 3906/15625
C = 1 + 2 + 4 + 8 + 16 + ... + 2048 + 4096
2C = 2 + 4 + 8 + 16 + 32 + ... + 4096 + 8192
2C - C = (2 + 4 + 8 + 16 + 32 + ... + 4096 + 8192) - (1 + 2 + 4 + 8 + ... + 2048 + 4096)
B = 8192 - 1
B = 8191
bài 6: TÍNH NHANH NÂNG CAO LỚP 5
A= 1/5+1/25+1/125+1/625+1/3125
giải hộ mình nhanh nhanh nha các bạn
5A=(1/5+1/25+1/125+1/625+1/3125)*5
5A=(1+1/5+1/25+1/125+1/625)
5A-A=(1+1/5+1/25+1/125+1/625)-(1/5+1/25+1/125+1/625+1/3125)=1-1/3125
4A=1-1/3125=3124/3125
A=3124/3125:4=781/3125.
\(A\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)=\frac{8}{5}\)
\(A\times\frac{3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{4\times6}{5\times5}=\frac{8}{5}\)
\(A\times\frac{3\times2\times4\times3\times5\times4\times6}{2\times2\times3\times3\times4\times4\times5\times5}=\frac{8}{5}\)
\(A\times\frac{6}{2\times5}=\frac{8}{5}\)
\(A\times\frac{3}{5}=\frac{8}{5}\)
\(A=\frac{8}{5}:\frac{3}{5}\)
\(A=\frac{8}{3}\)
ta có:
A = 8/5 : [(1 - 1/4) x ... x (1 - 1/5)]
= 8/5 : [3/4 x 8/9 x 15/16 x 24/25]
= 8/5 : 3/5
= 8/5 x 5/3
= 8/3
... là như đề bài nha!
=> C = 3/4 . 8/9 . 15/16 . 24/25
=> C = 3.4.15.24/4.9.16.25
=> C = 1.2.3.6/1.2.4.5 = 3.6/4.5 = 18/20
Sửa lại dấu chia thành dấu nhân thì bài toán mới đúng!
\(C=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right)\)
\(C=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}\)
\(C=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}\)
\(C=\frac{1.2.3.4}{2.3.4.5}.\frac{3.4.5.6}{2.3.4.5}=\frac{1}{5}.\frac{6}{2}=\frac{3}{5}\)
a.1/4 +3/8 +5/16=15/16
b, 3/5 - 1/3 -1/6=1/10
c, 4/7 *5/8 *7/12=5/24
d. 25/28 : 15/24 * 6/7=60/49
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)
\(=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}\)
\(=\frac{2}{6}\)
\(=\frac{1}{3}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1}{5}\)
\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}.\frac{8}{9}....\frac{35}{36}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}.\frac{5.7}{6.6}\)
\(=\frac{7}{2.6}=\frac{7}{12}\)
P/S: dấu "." là dấu nhân nhé
A = 1/4 +1/9 + 1/16 + 1/25 + 1/36
= ( 1/4 + 1/16 ) + ( 1/9 + 1/36) + 1/25
= 5/16 + 5/36 + 1/25
= 65/144 + 1/25
= 1769/3600
=> 1769/3600 < 5/6 (hay 1769/3600 < 3000/3600 -quyđồng-)
Vậy A< 5/6
Đúng nhé, tk cho mjk với-số to thiệt nhưng đúng mà-
\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)
=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)
=>\(A\cdot3=8\)
=>A=8/3
a: \(4C=4+1+...+\dfrac{1}{1024}\)
=>\(3C=4-\dfrac{1}{4096}=\dfrac{16383}{4096}\)
hay \(C=\dfrac{5461}{4096}\)
b: \(\dfrac{1}{5}D=\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{5^6}\)
=>\(\dfrac{-4}{5}\cdot D=\dfrac{1}{5^6}-\dfrac{1}{5}=\dfrac{1-5^5}{5^6}\)
=>\(D=\dfrac{5^5-1}{5^6}:\dfrac{4}{5}=\dfrac{5^5-1}{4\cdot5^5}\)