\(B=\dfrac{5,8}{\left|2,5-x\right|+5,8}\)

\(\left|2,5-x\right|\ge0\Rightarrow\left|2,5-x\right|+5,8\ge5,8\\ \Rightarrow B\le1\)

Max B =1 khi x=2,5

Đề bài là gì ạ?

\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)

Mà \(\left|4x-3\right|\ge0\)với mọi x

\(\left|5y+7,5\right|\ge0\)với mọi y

\(\Rightarrow A\)có GTNN là 10

Để A có GTNN thì :

\(4x-3=0\)                           \(5y+7,5=0\)

\(4x=3\)                                                  \(5y=-7,5\)

\(x=\frac{3}{4}\)                                                     \(y=-1,5\)

\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)

Mà \(\left|2,5-x\right|\ge0\)

\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)

Để B có GTLN \(\Rightarrow2,5-x=0\)

\(\Rightarrow x=2,5\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{3n+2-2}{2\left(3n+2\right)}=\dfrac{n}{2\left(3n+2\right)}\)

Bn viết để sai rồi, mk sửa lại :)

\(S\left(n\right)=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow S\left(n\right)=\dfrac{\dfrac{1}{2}-\dfrac{1}{3n+2}}{3}\)

Ta có: \(\left|2,5-x\right|\ge0;\forall x\)

\(\Rightarrow\left|2,5-x\right|+5,8\ge5,8;\forall x\)

\(\Rightarrow\frac{11,6}{\left|2,5-x\right|+5,8}\le2;\forall x\)

Dấu "="xảy ra \(\Leftrightarrow\left|2,5-x\right|=0\)

                        \(\Leftrightarrow x=2,5\)

Vậy\(M_{max}=2\)\(\Leftrightarrow x=2,5\)