tính
a; A = 3/4 * 8/9 * 15/16 * ..............* 9999/10000
b; B = {1 - 1/21 } * { 1 - 1/28 } * {1 - 1/36 } * .......................*{ 1 - 1326}
c; C = { 1 + 1/1*3 } * { 1 + 1 / 2*4 } * { 1 + 1/3*5} * ...........................* { 1+ 1/99*101}
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a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
a) ta có: A = 2156 + 357
=> A - 357 = 2156 + (357-357)
A - 357 = 2156
=> A - 2156 = (2156-2156) + 357
A - 2156 = 357
b) ta có: B = 9475 - 7436
=> B + 7436 = 9475 - (7436-7436)
B + 7236 = 9475
=> 9475 - B = 9475 + 9475 - 7436
9475 -B = 2.9475 - 7436
( phần b câu 2 hình như bn chép sai)
a)A - 357 = 2156 và A - 2156 = 357
b)B + 7436 = 9475 và 9475 - B = 7436
1)
a) \(34,2:36=0,95\)
b) \(36,18:18=2,01\)
c) \(5,05:25=0,202\)
d) \(17,25:75=0,23\)
2)
a) \(75,52:32=2,36\)
b) \(4,68:13=0,36\)
c) \(81,6:34=2,4\)
d) \(134,4:42=3,2\)
3)
a) \(882:36=24,5\)
b) \(78:24=3,25\)
c) \(3:15=0,2\)
d) \(15:75=0,2\)
4)
a) \(3:75=0,04\)
b) \(750:300=2,5\)
c) \(390:12=32,5\)
d) \(54:24=2,25\)
b) So hang thu 100 la :
3 + ( 100 - 1 ) x 2 = 201
( mk ko chac dau nhung chac la dung day, con may cau kia thi co nguoi giai roi )
Số số hạng là:
(701-3):2+1=350(số)
Số hạng thứ 100 là:
3+99 x 2=201
Tổng là:
(701+3) x 350:2=123200
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)
\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)