\(\frac{x+1}{97}\) + \(\frac{x+1}{98}\) - \(\frac{x+1}{99}\) - \(\frac{x+1}{100}\) \(\Leftrightarrow\) (x+1).(1/97 + 1/98 - 1/99 - 1/100) . Vì (1/97 = 1/ 98 - 1/99 - 1/100) \(\ne\) 0 \(\Rightarrow\) x+ 1= 0 \(\Leftrightarrow\) x= -1

1+1=3 :)))

a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)

\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)

\(=\frac{1}{100}\)

b) Tương tự như câu a

Ra kết quả phần b là \(\frac{101}{2}\) đúng hông?

giải hộ mik với min hf rất cần

Ta có : 

\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}=0\)

\(\Leftrightarrow\)\(\frac{x-100}{99}-\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}-\frac{x-100}{95}-\frac{x-100}{94}=0\)

\(\Leftrightarrow\)\(\left(x-100\right)\left(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)

Vì \(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)

Nên \(x-100=0\)

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

Vậy \(x=100\)

Bài làm mang tính chất tham khảo vì em mới lớp 7 ~

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{k}{x\left(x+100\right)}\)

\(\frac{1}{x}-\frac{1}{x+100}=\frac{k}{x\left(x+100\right)}\)

\(\frac{x+100}{x\left(x+100\right)}-\frac{x}{x\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)

k = 100

k=100

1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

=> x = - 1

b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)

=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)

=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

=> x + 2010 = 0

=> x = -2010

c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)

=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)

=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)

=> x = -1900

d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)

=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)

=> x = -2028

1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

        \(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

        \(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

  + TH₁: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

  + TH₂: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

            \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

             mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

             \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1\)

2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

        \(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)

        \(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)

        \(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

  + TH₁: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)

  + TH₂: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)

              \(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)

               mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2010\)

3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

        \(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)

        \(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

       \(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

  + TH₁: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)

  + TH₂: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)

              \(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)

               mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1900\)

4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

         \(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)

         \(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

         \(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

  + TH₁: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)

  + TH₂: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)

              \(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)

               mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=104\)

5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

        \(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)

        \(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

        \(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

    + TH₁: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)

    + TH₂: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)

              \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)

               mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2028\)

Chúc bn hok tốt nha

đợi mk 1 chút nha

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\left(tm\right)\)

Cho mình sửa lại đề nhá:Chỉ có 1 cái \(\frac{1}{2}x^{100}\)thôi.Xin lỗi