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25 tháng 9 2016

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

25 tháng 9 2016

x thuoc R

27 tháng 3 2018

Không biết

28 tháng 11 2016

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

7 tháng 10 2016

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

=> \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

=> \(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

=> \(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

=> \(x+99=0\) (Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\) )

=>\(x=-99\)

9 tháng 10 2016

Ta có :

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

\(\Rightarrow\)  \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

\(\Rightarrow\left(x+99\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Vì  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\)

\(\Rightarrow\)  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\ne0\)

Nên  \(x+99=0\)

\(\Rightarrow x=0-99\)

\(\Rightarrow x=-99\)

Vậy :        \(x=-99\)

28 tháng 9 2017

a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)

\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)

Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

Vậy x = -5

b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)

\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)

\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)

\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

\(\Rightarrow x+102=0\Rightarrow x=-102\)

Vậy x = -102

c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3

                                      = x - x + 2 - 3

                                      = -1

mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0

d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)

\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)

\(\Rightarrow x\ge\frac{-7}{3}\)

Vậy \(x\ge\frac{-7}{3}\)

24 tháng 6 2018

\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)

\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

Nên \(x+100=0\)

\(\Rightarrow\)\(x=-100\)

Vậy \(x=-100\)

Chúc bạn học tốt ~ 

24 tháng 6 2018

\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)

\(\Leftrightarrow\)\(x+1=2009\)

\(\Leftrightarrow\)\(x=2009-1\)

\(\Leftrightarrow\)\(x=2008\)

Vậy \(x=2008\)

Chúc bạn học tốt ~ 

20 tháng 7 2016

a/ \(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\)     (ĐK: \(x\ne\frac{3}{2};x\ne-2\))

     \(\Rightarrow\left(x+2\right)\left(5x-4\right)=\left(7+4x\right)\left(3-2x\right)\)

     \(\Rightarrow5x^2-4x+10x-8=21-14x+12x-8x^2\)

     \(\Rightarrow13x^2+8x-29=0\)

      \(\Rightarrow13\left(x^2+\frac{8}{13}x-\frac{29}{13}\right)=0\)

     \(\Rightarrow13\left[x^2+2.\frac{4}{13}.x+\left(\frac{4}{13}\right)^2-\left(\frac{4}{13}\right)^2-\frac{29}{13}\right]=0\)

      \(\Rightarrow13\left[\left(x+\frac{4}{13}\right)^2-\frac{393}{169}\right]=0\)

       \(\Rightarrow13\left(x+\frac{4}{13}\right)^2-\frac{393}{13}=0\)

        \(\Rightarrow\left(x+\frac{4}{13}\right)^2=\frac{393}{169}\)

       \(\Rightarrow\orbr{\begin{cases}x+\frac{4}{13}=\sqrt{\frac{393}{169}}=\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4+\sqrt{393}}{13}\\x+\frac{4}{3}=-\sqrt{\frac{393}{169}}=-\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4-\sqrt{393}}{13}\end{cases}}\)

        Vậy biểu thức có 2 nghiệm \(x=\left\{\frac{-4+\sqrt{393}}{13};\frac{-4-\sqrt{393}}{13}\right\}\)

b/ \(\frac{x-1}{99}+\frac{x-2}{98}-\frac{x-3}{97}-\frac{x-4}{96}=0\)

   \(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1-\left(\frac{x-3}{97}-1\right)-\left(\frac{x-4}{96}-1\right)=0\)

   \(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}=0\)

   \(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

    => x - 100 = 0 => x = 100

                                                    Vậy x = 100

28 tháng 9 2017

a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)

\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)

\(\Rightarrow\)\(x+102=0\)

\(\Rightarrow x=-102\)

28 tháng 9 2017

c) \(\left(x+2\right)-\left(x+3\right)>0\)

\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)

\(\Rightarrow x\in\varnothing\)

d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)

TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\ge\dfrac{-7}{3}\)

TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\le5\)

TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)

26 tháng 8 2018

a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0

=> x + 100 = 0 => x = -100

b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)

\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)

\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)

Vì 1/47 + 1/48 - 1/49 khác 0

Nên x -50 = 0 => x = 50