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3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
4)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
2, đặt tên biểu thức trên là A. Ta có :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
TÍNH NHANH
B=2 phần 1.3 + 2 phần 3.5 + 2 phần 5.7 +................+ 2 phần 99.101
(Giải thích rõ nha)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+..........+\frac{2}{99.101}\)
B=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...........+\frac{1}{99}-\frac{1}{101}\)
B=\(1-\frac{1}{101}\)
B=\(\frac{100}{101}\)
Gọi số số hạng vế trái của đẳng thức là : m(m ∈ N*)
Ta có: (11+x-3).m : 2= 0
(11+x-3).m=0
Mà m ∈ N*=> m ≠ 0
=> 11+x-3=0
=> 11+x =0+3
=> 11+x=3
=> x=3-11
=>x= -8
Dễ hiểu nhất rùi đó nha!!!
\(S_n=1.1!+2.2!+3.3!+...+n.n!\)
\(\text{Ta có:}\) \(1.1!=2!-1!\)
\(2.2!=3!-2!\)
\(3.3!=4!-3!\)
.......
\(n.n!=\left(n+1\right)!-n!\)
Cộng vế với vế ta đc:
\(S_n=1.1!+2.2!+3.3!+...+n.n!=2!-1!+3!-2!+4!-3!+...+\left(n+1\right)!-n!\)
\(=\left(n+1\right)!-1!=\left(n+1\right)!-1\)
1 +( -2) + 3 + (-4) +...+2001 + (-2002) + 2003
= [1 +( -2)] + [3 + (-4)] +...+ [-2000+2001] + [(-2002) + 2003]
= -1 + -1 +............ + 1 + 1
= 0
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+.....+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{8}{9}+\frac{1}{2}\cdot\frac{2}{5}=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
=1/2x(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.....+1/8-1/10)
=1/2x58/45
=29/45
a) 5x - x = 64 \(\Rightarrow\) 4x = 64 \(\Rightarrow\) x = 16
b) \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
c) \(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
d) \(C=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)