K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

sai đề                                                            

23 tháng 6 2015

\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=3\)

x+1/1974+1+x+2/1973+1+x+3/1972=0

x+1+1974/1974+x+2+1973/1973+x+3+1972/1972=0

x+1975/1974+x+1975/1973+x+1975/1972=0

(x+1975)(1/1974+1/1973+1/1972)=0

=>x+1975=0 vì 1/1974+1/1973+1/1972 Khác 0

=>x=-1975

27 tháng 6 2015

\(\frac{x+1}{1974}+1+\frac{x+2}{1973}+1+\frac{x+3}{1972}+1=-3+3\)

\(\Leftrightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)

\(\Leftrightarrow\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)

=> x  + 1975 = 0. => x = -1975 ( vì 1/1974 + 1/1973 + 1/1972 khác 0 )

23 tháng 7 2015

\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)

\(\Rightarrow\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=0\)

\(\Rightarrow\frac{x+1+1974}{1974}+\frac{x+2+1973}{1973}+\frac{x+3+1972}{1972}=0\)

\(\Rightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)

\(\Rightarrow\left(x+1975\right)\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}=0\)

Mà \(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\ne0\)

 \(\Rightarrow x+1975=0\)

\(\Rightarrow x=0+1975\)

\(\Rightarrow x=1975\)

Vậy \(x=1975\)

b) phần này làm tương tự phần a nha, chuyển -3 sang vế bên trái r cộng từng p.số vs 1 và sau đó nhóm tử số chung ra ngoài ^^

22 tháng 7 2019

\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)

\(=>\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=-3+3\)

\(=>\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)

\(\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)

\(=>x+1975=0=>x=-1975\)

Vậy \(x=-1975\)

22 tháng 7 2019

\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)

\(\Leftrightarrow\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=0\)

\(\Leftrightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)

\(\Leftrightarrow\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)

\(\Leftrightarrow x+1975=0\)

\(\Leftrightarrow x=-1975\)

15 tháng 7 2018

a) \(\frac{2}{7}x-\frac{1}{3}x=\frac{5}{21}\)

\(\left(\frac{2}{7}-\frac{1}{3}\right)x=\frac{5}{21}\)

\(\left(-\frac{1}{21}\right)x=\frac{5}{21}\Rightarrow x=\frac{5}{21}:-\frac{1}{21}=-5\)

b) \(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)

\(\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=-3+3\)

\(\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)

\(\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)

Mà \(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}>0\Rightarrow x+1975=0\)

\(x=-1975\)

15 tháng 7 2018

a) x = - 5

b) x= - 1975

28 tháng 9 2016

d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)

<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)

<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)

<=> x = -2010

28 tháng 9 2016

Làm câu khó nhất rồi, còn lại tự làm nha <(") /_\

5 tháng 6 2019

1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)

\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu

\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)

\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)

Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)

5 tháng 6 2019

Bài 1b) có thể giải gọn hơn nhuư thế này

14 tháng 8 2020

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

14 tháng 8 2020

cảm ơn