Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left[x+1\right]}=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{\frac{2017}{2019}}{2}=\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}=\frac{1}{2019}\)
=> x + 1 = 2019 <=> x = 2018
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Leftrightarrow x=308-3\)
\(\Leftrightarrow x=305\)
Vậy \(x=305\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x-1\right)}=\)\(\frac{2017}{2019}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x-1\right)}=\frac{2017}{2019}\)
\(2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right]=\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\)\(\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2019}\)
x + 1 =2019
x = 2019-1 =2018
Vậy x = 2018
\(2\left(\frac{1}{3}.\frac{1}{2}+\frac{1}{6}.\frac{1}{2}+\frac{1}{10}.\frac{1}{2}+....+\frac{2}{x\left(x+1\right)}.\frac{1}{2}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2[\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+....+\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+1}]=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}+0+0+....+0-\frac{1}{x-1}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=>\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{4038}\)
=>\(\frac{1}{x+1}=\frac{1}{2019}\)
=> x+1=2019
=>x=2018
\(1.\left(x+3\right)^3=\frac{1}{-27}\)
\(\left(x+3\right)^3=\left(\frac{1}{-3}\right)^3\)
\(\Rightarrow x+3=\frac{1}{-3}\)
\(\Rightarrow x=\frac{-1}{3}-3\)
\(x=\frac{-10}{3}\)
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
Tim x biet
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)
=> \(1-\frac{2}{x+1}=\frac{2019}{2020}\)
=> \(\frac{2}{x+1}=\frac{1}{2020}=\frac{2}{4040}\)
=> x + 1 = 4040 => x = 4039
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2015}{2017}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2015}{2017}\\ 2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2015}{2017}\\ \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2015}{2017}:2\\ \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2015}{4034}\\ \dfrac{1}{x+1}=\dfrac{1}{2017}\\ \Rightarrow x+1=2017\\ x=2016\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{2019}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2019}\)
\(\Leftrightarrow x+1=2019\)
\(\Leftrightarrow x=2019-1\)
\(\Leftrightarrow x=2018\)
Vậy x = 2018
chị ơi bài này em học từ lớp 6 rồi( ngay đầu vào luôn ) nhung dốt nát lè