Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Khi đó:
$1+\frac{1}{1.3}=\frac{2^2}{1.3}$
$1+\frac{1}{2.4}=\frac{3^2}{2.4}$
.........
$1+\frac{1}{99.101}=\frac{100^2}{99.101}$
Khi đó:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
\(x\times100567=1467171963\)
\(x=1467171963\div100567\)
\(x=14589\)
Vậy x=14589
\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{49.51}\right)\)+\(\dfrac{2}{51}\)
=\(\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{2500}{49.51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{50^2}{49.51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{\left(2.3.4.....50\right)\left(2.3.4.....50\right)}{\left(1.2.3.....49\right)\left(3.4.....51\right)}\)+\(\dfrac{2}{51}\)
=\(\dfrac{\left(2.3.4.....49\right).50.2.\left(3.4.5.....50\right)}{1.\left(2.3.4.....49\right)\left(3.4.5.....50\right).51}\)+\(\dfrac{2}{51}\)
=\(\dfrac{50.2}{1.51}\)+\(\dfrac{2}{51}\)=\(\dfrac{100}{51}\)+\(\dfrac{2}{51}\)=\(\dfrac{102}{51}\)=2
1) 2x = 1- 3
2x = -2
x = -2 : 2
x = -1
2) 2x = 5 - 3
2x = 2
x = 2 : 2
x = 1
\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)
\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)
\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)
\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)
\(A=\frac{100\cdot2}{1\cdot101}\)
\(A=\frac{200}{101}\)