Khoảng cách có rồi thì bạn áp dụng công thức : \(\frac{a}{m.n}=\frac{1}{m}-\frac{1}{n}\)(với n-m=a) là làm được

S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{93.95}+\frac{3}{95.98}+\frac{4}{98.102}+\frac{5}{102.17}+\frac{2012}{107..2119}\)

S=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+\frac{1}{102}-\frac{1}{107}+\frac{1}{107}-\frac{1}{2119}\)

S=\(\frac{1}{5}-\frac{1}{2119}\)

S=\(\frac{2114}{10595}\)

4/3.5+4/5.7+4/7.9+4/9.11

=4.(1/3.5+1/5.7+1/7.9+1/9.11)

=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)

=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)

=2.(1/3-1/11)

=2.8/33

=16/33

  4/3.5+4/5.7+4/7.9+4/9.11

=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11

=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11

=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)

=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)

=2.(1/3-1/11)

=2.8/33

=16/33

16/33 nha

Ta có: \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.3^2^{ }=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.9=351\)

\(\Rightarrow3^x.\left(1+3+9\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=351:13=27\)

\(\Rightarrow x=3\)

3 k mk nha

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3^x+3^x+1+3^x+2=351

3^x+3^x.3+3^x.3²=351

3^x.(1+3+3²)=351

3^x.(1+3+9)=351

3^x.13=351

3^x=351:13

3^x=27

3^x=3³

  x=3

Vậy x=3

\(\Leftrightarrow-\frac{1}{6}< -\frac{1}{3}x+2< \frac{1}{6}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{3}x+2>-\frac{1}{6}\\-\frac{1}{3}x+2< \frac{1}{6}\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{13}{2}\\x>\frac{11}{2}\end{cases}\Leftrightarrow\frac{11}{2}< x< \frac{13}{2}}\)

vậy

Xét 2 Th nha :

 Th1 : \(\left|-\frac{1}{3}x+2\right|< 0\)

PT trở thành : \(\frac{1}{3}x-2< \frac{1}{6}\)

\(\Rightarrow\frac{1}{3}x< \frac{13}{6}\)

\(\Rightarrow x< \frac{13}{2}\)

Th2 : \(\left|-\frac{1}{3}x+2\right|\ge0\)

\(\Rightarrow\frac{-1}{3}x+2< \frac{1}{6}\)

\(\Rightarrow\frac{-1}{3}x< \frac{-11}{6}\)

\(\Rightarrow x>\frac{11}{2}\)

Tự kết luận nha . Nhớ xét điều kiện nha

\(\left|-\frac{1}{3}x+2\right|\)luôn dương mà

a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20

b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)

      =>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008

Ta có: \(3\left|x^2-1\right|-6=\left|1-x^2\right|\)

\(\Leftrightarrow3\left|x^2-1\right|-\left|x^2-1\right|=6\)

\(\Leftrightarrow2\left|x^2-1\right|=6\)

\(\Leftrightarrow\left|x^2-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=3\\x^2-1=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=-2\end{cases}}\)

Vì \(x\ge0>-2\left(\forall x\right)\)

\(\Rightarrow x^2=4\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)