tính
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+49+50}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 5 2022
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
16 tháng 3 2018
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
21 tháng 3 2017
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
20 tháng 7 2022
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
5 tháng 5 2018
P = 1/49+2/48+3/47+...+48/2+49/1
Cộng 1 váo mỗi p/s trong 48 p/s đầu , trừ p/s cuối đi 48 ta được
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50
Đưa ps cuối lên đầu
P=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50S
=> S/P=1/50
S
10 tháng 3 2019
\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)
18 tháng 7 2018
Ta có:
P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\left(1+1+...+1\right)\)(có 49 chữ số 1)
P= \(\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
P= \(\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
P= \(50.\left(\dfrac{1}{50}+\dfrac{1}{49}+...+\dfrac{1}{2}\right)\)
⇒ \(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}}{50.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)}\)
⇒ \(\dfrac{S}{P}=\dfrac{1}{50}\)
Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)
27 tháng 7 2018
Ta có :
Vế phải =1 - 1/2 + 1/3 - 1/4 + ... + 1/49 - 1/50
= (1+ 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + ... +1/50)
<=> (1 + 1/2 + 1/3 + 1/4 + ... + 1/49+1/50)- 2(1/2 +1/4 +...+1/50)
=(1+1/2 +1/3 +1/4...+ 1/49+1/50) - (1+1/2 +...+1/25)
=1/26 + 1/27 +1/28 +...+1/50 (đpcm)
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+49+50}\)
\(=\dfrac{1}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3\left(3+1\right)}{2}}+\dfrac{1}{\dfrac{4\left(4+1\right)}{2}}+...+\dfrac{1}{\dfrac{50\left(50+1\right)}{2}}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{51}\right).2\)
\(=\dfrac{49}{102}.2\)
\(=\dfrac{49}{51}\)
Lời giải:
Sử dụng công thức:
\(1+2+....+n=\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{1}{1+2+3+...+n}=\frac{2}{n(n+1)}\)
Do đó:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)
\(\Rightarrow \frac{A}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{51-50}{50.51}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}=\frac{1}{2}-\frac{1}{51}\)
\(\Rightarrow A=1-\frac{2}{51}=\frac{49}{51}\)