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23 tháng 7 2017

a. \(\frac{x-5}{2000}+\frac{x-4}{1999}+\frac{x-3}{1998}=\frac{x-2}{1997}+\frac{x-1}{1996}+\frac{x}{1995}\)

\(\Leftrightarrow\left(\frac{x-5}{2000}+1\right)+\left(\frac{x-4}{1999}+1\right)+\left(\frac{x-3}{1998}+1\right)=\left(\frac{x-2}{1997}+1\right)+\left(\frac{x-1}{1996}+1\right)+\left(\frac{x}{1995}+1\right)\)

\(\Leftrightarrow\left(x+1995\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}-\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)=0\)

\(\Leftrightarrow x+1995=0\)

\(\Leftrightarrow x=-1995\)

23 tháng 7 2017

CÂU B BẠN LÀM TƯƠNG TỰ NHÉ

2 tháng 10 2017

\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)

\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

\(=>x+2001=0\)

\(x=-2001\)

\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)

\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(=>x+1998=0\)

\(x=-1998\)

6 tháng 4 2018

dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

x=-2000           

3 tháng 8 2015

ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)

        \(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)

     \(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)  (1)

                     Xét     \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)

                                        từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )

                                                          \(\Rightarrow x=2000\)

5 tháng 4 2016

a/  (X+1)/35+1+(x+3)/33+1 =(x+5)/31+(x+7)/29+1+1

=>(x+36)/35+(x+36)/33-(x+36)/31-(x+36)/27=0

=>(X+36)(1/35+1/33-1/31-1/29)=0

=> x+36=0(vì c=vế 2 luôn luôn khác 0)

=>x=-36

b/ CMTT câu a 

trừ tung phân số cho 1 ta được x=2004

5 tháng 4 2016

Ngu người khi ko biết làm bài lày

12 tháng 3 2020

a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

12 tháng 3 2020

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999

27 tháng 2 2020

Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)

\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)

\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)

\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)

\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)

\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)

Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)

=> x - 2000 = 0 

=> x = 2000

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

13 tháng 8 2019

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010

18 tháng 3 2020

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

18 tháng 3 2020

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)

8 tháng 7 2017

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

1 tháng 8 2017

khó lắm

bây h thì bạn giải đc chưa

25 tháng 2 2020

\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)

suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)

suy ra  \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)\(\frac{x-2004}{2}\)\(\frac{x-2004}{4}\)\(\frac{x-2004}{6}\)\(\frac{x-2004}{8}\)\(\frac{x-2004}{10}\)=0

suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)\(\frac{1}{8}\)\(\frac{1}{10}\))=0

Vì  \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)\(\frac{1}{8}\)\(\frac{1}{10}\) khác 0

nên x-2004=0 suy ra x=2004

27 tháng 2 2020

em cảm ơn