K
Khách

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

10 tháng 5 2018

a,\(\Leftrightarrow\left(\frac{1-x}{2013}+1\right)=\left(\frac{2-x}{2012}+1\right)-\left(1-\frac{x}{2014}\right)\)

   \(\Leftrightarrow\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{2014-x}{2014}\)

   \(\Leftrightarrow\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0

   \(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\right)=0\)

   \(\Leftrightarrow x=2014\left(do.cái.còn.lại.\ne0\right)\)

b,tương tự +1 vào cái thứ nhất ,+1 vào cái thứ 2,1- vào cái thứ 3 được x=2013

10 tháng 5 2018

ban oi them bot sai roi

\(c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.\)

\(\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0\)

\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

=>\(\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}\)

Vậy.......

3 tháng 2 2020

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2001}+1\right)+\left(\frac{-x}{2003}+1\right)\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow\left(2003-x\right)=0\) (vì \(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))

\(\Leftrightarrow x=2003\).

Vậy tập nghiệm của phương trình là \(S=\left\{2003\right\}\).

6 tháng 7 2019

a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

b) Sửa đề :

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow x=300\)

c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)

\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)

\(\Leftrightarrow x=2004\)

Vậy....

11 tháng 1 2017

Theo bài ra , ta có :

\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)

\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)

\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)

\(\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

Vậy \(x=-2016\)

Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)

Chúc bạn học tốt =)) 

11 tháng 1 2017

\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)

\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)

\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)

\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

11 tháng 2 2020

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)

Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0

Vậy x = 2012

11 tháng 2 2020

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10

2 tháng 3 2018

2.

pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0

<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0

<=> (x-2000).(1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004) = 0

<=> x-2000=0 ( vì 1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004 > 0 )

<=> x=2000

Tk mk nha

2 tháng 3 2018

1.

a, = (2x-1)^2-2.(2x-1)+1-4

    = (2x-1-1)^2-4

    = (2x-2)^2-4

    = (2x-2-2).(2x-2+2)

    = 2x.(2x-4)

b, = [x.(x+3)].[(x+1).(x+2)]

    = (x^2+3x).(x^2+3x+1)-8

    = (x^2+3x+1)^2-1-8

    = (x^2+3x+1)^2-9

    = (x^2+3x+1-3).(x^2+3x+1+3)

    = (x^2+3x-2).(x^2+3x+4)

    = ((x+1).(x+3).(x^2+3x-2)

Tk mk nha