1 .200
Tính A=(200-1).(200-2).(200-3)...
Có 200 thừa số
Vì A có 200 thừa số nên thừa số cuối cũng sẽ là ( 200 - 200 )
Khi đó :
A = ( 200 - 1 ) ( 200 - 2 ) ... ( 200 - 200 )
A = ( 200 - 1 ) ( 200 - 1 ) ... 0
A = 0
Vậy A = 0
A=0 nha
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hok tốt
hãy so sánh 20020 -2/20019- 2 và 20019 +1/20018+ 1
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)
=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\)
=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)
2.
E=1+1/2.(1+2)+1/3.(1+2+3)+...+1/200.(1+2+...+200)
\(E=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{\left(201-2+1\right)\cdot\left(201+2\right)}{4}=\dfrac{200\cdot203}{4}=50\cdot203=10150\)
Cho A = 1 x 200 + 2 x 199 + 3 x 198 + ... 200 x 1 và B = 1 + (1 + 2) + (1 + 2 + 3 + ... + 200). Tính A - B
Tính A - B
là dekisugi thông minh mà sao lại phải đi hỏi thế
Cho
A = 1 x 200 + 2 x 199 + 3 x 198 + ... 200 x 1 và B = 1 + (1 + 2) + (1 + 2 + 3 + ... + 200).
Tính A - B
Tính A - B
200*200 + 1 =
CMR: 1-1/2+1/3-1/4+....+1/199-1/200=1/101-1/102+...+1/200
CMR :: 1-1/2+1/3-1/4+.....+1/199-1/200 =1/101+1/102+...+1/200
Chứng minh rằng :1-1\2+1\3+...+1\999+1\200=1\101+1\102+...+1\200
Sửa đề: \(CMR:1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{3}+...+\frac{1}{199}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)(ĐPCM)