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\(\left(1-\frac{1}{1931}\right)\left(1-\frac{1}{1932}\right)......\left(1-\frac{1}{2012}\right)\)
\(=\frac{1930}{1931}\cdot\frac{1931}{1932}\cdot...\cdot\frac{2011}{2012}\)
\(=\frac{1930\cdot1931\cdot...\cdot2011}{1931\cdot1932\cdot...\cdot2012}=\frac{1930}{2012}=\frac{965}{1006}\)
\(\left(1-\frac{1}{1931}\right)\times\left(1-\frac{1}{1932}\right)\times\left(1-\frac{1}{1933}\right)\times...\times\left(1-\frac{1}{2012}\right)\)
\(=\left(\frac{1931}{1931}-\frac{1}{1931}\right)\times\left(\frac{1932}{1932}-\frac{1}{1932}\right)\times\left(\frac{1933}{1933}-\frac{1}{1933}\right)\times...\times\left(\frac{2012}{2012}-\frac{1}{2012}\right)\)
\(=\frac{1930}{1931}\times\frac{1931}{1932}\times\frac{1932}{1933}\times...\times\frac{2011}{2012}\)
\(=\frac{1930\times1931\times1932\times...\times2011}{1931\times1932\times1933\times...\times2012}\)
\(=\frac{1930}{2012}=\frac{965}{1006}\)
1) So sánh
3 77/379 và 3 79/381
2)
A= 1/6 + 1/10 + 1/15 + 1/21 + 1/28 + 1/36
Giúp mình nhé❤❤❤❤❄▫〰▫▫▫▫▫▫
2) A = \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{2}\).\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\right)\)
=> \(\frac{1}{2}\).A = \(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{9}\)
=> \(\frac{1}{2}\).A = \(\frac{2}{9}\)
=> A = \(\frac{2}{9}:\frac{1}{2}\)
=> A = \(\frac{4}{9}\)
Bài 1 :
A = 12 + 22 + 32 +....+n2
A = 12 + 2.(1+1) + 3.(2 +1) + 4.( 3 +1) +.....+n(n-1 + 1)
A = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 +.....+ n.(n-1) + n
A = ( 1 + 2 + 3 + 4 +....+n) + ( 1.2 + 2.3 + 3.4 +....+(n-1).n
A = (n+1).{(n-1):n+1)/2 +1/3.[1.2.3 +2.3.3 +.....+(n-1)n.3]
A = (n+1).n/2+1/3.[1.2.3 +2.3.(4-1)+ ...+(n-1).n [(n+1) - (n -2)]
A = (n+1)n/2+1/3.( 1.2.3 + 2.3.4 -1.2.3 +..+ (n-1)n(n+1)- (n-2)(n-1)n)
A =(n+1)n/2 + 1/3.(n-1)n(n+1)
A = n(n+1)[1/2 + 1/3 .(n-1)]
A = n.(n+1) \(\dfrac{3+2n-2}{6}\)
A= n.(n+1)(2n+1)/6
Bài 2 :
a, (x+1) +(x+2) + (x+3)+...+(x+10) = 5070
(x+10 +x+1).{( x+10 - x -1): 1 +1):2 = 5070
(2x + 11)10 : 2 = 5070
( 2x + 11)5 = 5070
2x+ 11 = 5070:5
2x = 1014 - 11
2x = 1003
x = 1003 :2
x = 501,5
b, 1 + 2 + 3 +...+x = 820
( x + 1)[ (x-1):1 +1] : 2 = 820
(x +1).x = 820 x 2
(x +1).x = 1640
(x +1) .x = 40 x 41
x = 40
mình chỉ giúp bạn phần a thôi nhé, còn lại bạn tự suy nghĩ
gọi d=( n+1, 2n+1)
=> n+1 chia hết cho d=> 2n+2 chia hết cho d
=>2n+1 chia hết cho d=> 2n+1 chia hết cho d
=> ( 2n+2)-( 2n+1) chia hết cho d
=> 1 chia hết cho d
=> d= -1 hoặc +1
=> phân số n+1/2n+1 là phân số tối giản
\(x+\frac{2}{15}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{2}{15}\)
\(x=\frac{1}{5}\)
h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)
\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}:x=\frac{1}{12}\)
\(x=\frac{2}{3}:\frac{1}{12}\)
\(x=8\)
Ta có: \(\frac{1}{101}\)>\(\frac{1}{200}\)
\(\frac{1}{102}\)>\(\frac{1}{200}\)
\(\frac{1}{103}\)>\(\frac{1}{200}\)
...
\(\frac{1}{200}\)=\(\frac{1}{200}\)
=>A>\(\frac{1}{200}\).100 =>A>\(\frac{1}{2}\)
\(C=\left(1-\frac{1}{1931}\right)\left(1-\frac{1}{1932}\right)\left(1-\frac{1}{1933}\right).....\left(1-\frac{1}{2019}\right)\)
\(C=\frac{1930}{1931}\cdot\frac{1931}{1932}\cdot\frac{1932}{1933}\cdot\cdot\cdot\cdot\cdot\frac{2018}{2019}\)
\(C=\frac{1930\cdot1931\cdot1932\cdot\cdot\cdot\cdot\cdot2018}{1931\cdot1932\cdot1933\cdot\cdot\cdot\cdot\cdot2019}=\frac{1930}{2019}\)
(1-1/1931).(1-1/1932).(1-1/1933).........(1-1/2019)
= (1930/1931).(1931/1932).(1932/1933)......(2018/(2019)
= 1930/2019
@ Hc tốt nha cj !!!!