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BT
0
MH
31 tháng 12 2015
\(\Rightarrow A=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+\frac{2}{\left(4+1\right).4}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
Vậy A=1.
31 tháng 12 2015
Cái này có trong violympic vòng 10..bạn nhớ ôn cho kĩ nếu như bạn thi violympic!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 5 2016
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
NV
0
B
0
24 tháng 7 2018
(1/2+1/3+1/4+...+1/100)/(99/1+98/2+97/3+...+1/99)
=(1/2+1/3+1/4+...+1/100)/(1+100/2+100/3+100/4+....+100/99)
=(1/2+1/3+1/4+...+1/100)/100*(1/100+1/99+1/98+...+1/4+1/3+1/2)
=1/100
chỗ 99/1+99/2+99/3+...+1/99 hình như đề bài sai đề bài đúng hình như là trên đã sửa rồi
TN
0
19 tháng 9 2020
\(\text{Bài 4:}\)
\(a.\left|x-\frac{3}{5}\right|< \frac{1}{3}\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}>-\frac{1}{3}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x>\frac{4}{15}\end{cases}\Rightarrow\frac{4}{15}< x< \frac{14}{15}}\)
\(b.\left|-5,5\right|=5,5\)
\(\Rightarrow\left|x+\frac{11}{2}\right|>5,5\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>5,5\\x+\frac{11}{2}< -5,5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>0\\x< -11\end{cases}}\)
- Nhận xét:
Với mọi số dương n, ta có:
1 - n2 = ( 1 - n ) + ( n - n2 )
= ( 1 - n ) + n ( 1 - n )
= ( 1 - n )( 1 + n ).
Do đó:
\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\) \(\left(\frac{1}{99^2}-1\right)\)
= \(\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}.....\frac{1-99^2}{99^2}\)
= \(\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.\)\(\frac{\left(1-4\right)\left(1+4\right)}{4^2}.....\frac{\left(1-99\right)\left(1+99\right)}{99^2}\)
= \(\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}.....\frac{\left(-98\right).100}{99^2}\)
= \(\frac{\left(-1\right).3.\left(-2\right).4.\left(-3\right).5.....\left(-98\right).100}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(99.99\right)}\)
= \(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right).....\left(-98\right)\right]\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)
= \(\frac{\left(1.2.3.....98\right).\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)
= \(\frac{1.100}{99.2}\)
= \(\frac{50}{99}\).
\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{99^2}-1\right)\)
\(=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{9801}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9800}{9801}=\frac{-\left(1.3\right)}{2.2}.\frac{-\left(2.4\right)}{3.3}.\frac{-\left(3.5\right)}{4.4}...\frac{-\left(98.100\right)}{99.99}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}=\frac{1.3.2.4.3.5...97.99.98.100}{2.2.3.3.4.4...99.99}\)
\(=\frac{\left(1.2.3.4....98\right).\left(3.4.5.6...100\right)}{\left(2.3.4.5...99\right).\left(2.3.4.5...99\right)}=\frac{100}{99.2}=\frac{50}{99}\)