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DT
Dang Tung
CTVHS
21 tháng 6 2023
\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1}\\ =-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{48}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\dfrac{24}{25}\)
6 tháng 10 2018
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
9 tháng 4 2017
Bài 1:
Ta có:
\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)
Bài 2:
Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)
Nhận xét:
\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)
AV
8 tháng 5 2018
Ta có : A=(299+298+...+2+1-(249+248+...+2+1)250)/249+248+...+2+1
A= \(\dfrac{2^{99}+2^{98}+...+2+1-2^{99}-2^{98}-...-2^{51}-2^{50}}{2^{49}+2^{48}+...+2+1}\)
A=\(\dfrac{2^{49}+2^{48}+...+2+1}{2^{49}+2^{48}+...+2+1}\) = 1
Vậy đa thức A=1
NL
1
OK
0
NL
1
31 tháng 1 2022
a: \(=625\cdot3-4\cdot\left(15^4-1\right)\)
\(=1875-4\cdot50625+4\)
\(=1879-202500=-200621\)
b: =50+49+48+47+...+2+1
Số số hạng là 50-1+1=50(số)
Tổng của dãy số là:
\(\dfrac{\left(1+50\right)\cdot50}{2}=51\cdot25=1275\)
OK
0
OK
0
Đặt \(A=-2^{49}-2^{48}-...-2^1-1\)
\(\Rightarrow-A=2^{49}+2^{48}+...+2^1+1\\ \Rightarrow-2A=2^{50}+2^{49}+...+2^2+2^1\\ \Rightarrow-A-\left(-2A\right)=\left(2^{49}+2^{48}+...+2^1+1\right)-\left(2^{50}+2^{49}+...+2^2+2^1\right)\\ A=1-2^{50}\)
Thay vào \(2^{50}-2^{49}-2^{48}-...-2^1-1\) được:
\(2^{50}-2^{49}-2^{48}-...-2^1-1\\ =2^{50}+1-2^{50}\\ =1\)
`S = 2^50 -2^49 -2^48 -...-2^1 -1`
`2S = 2^51 - 2^50 - 2^49 - ... - 2^2 - 2`
`2S - S = (2^51 - 2^50 - 2^49 - ... - 2^2 - 2) - (2^50 -2^49 -2^48 -...-2^1 -1)`
`S = 2^51 - 2^50 - 2^49 - ... - 2^2 - 2 - 2^50 +2^49 +2^48 +...+2^1 +1`
`S = 2^51 - 2^50 - 2^50 + 1`
`S = 2^51 - (2^50 + 2^50) + 1`
`S = 2^51 - 2.2^50 + 1`
`S = 2^51 - 2^51 + 1`
`S = 1`