\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}+...+\dfrac{2}{8\times9\times10}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+\dfrac{1}{4\times5}-\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}-\dfrac{1}{9\times10}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{9\times10}\)

\(=\dfrac{1}{2}-\dfrac{1}{90}\)

\(=\dfrac{22}{45}\)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\left(x\ne0\right)\)

\(\Rightarrow x.\left(-x\right)=-2.\dfrac{8}{25}\)

\(\Rightarrow-x^2=-\dfrac{16}{25}\)

\(\Rightarrow x^2=\dfrac{16}{25}\)

\(\Rightarrow x^2=\left(\pm\dfrac{4}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

\(D=\dfrac{1}{2\left|x-1\right|+3}\)

Ta có:\(\left|x-1\right|\ge0\forall x\)

\(\Rightarrow2\left|x-1\right|\ge0\forall x\)

\(\Rightarrow2\left|x-1\right|+3\ge3\forall x\)

\(\Rightarrow\dfrac{1}{2\left|x-1\right|+3}\le\dfrac{1}{3}\forall x\)

Dấu "=" xảy ra  \(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy \(max\) \(E=\dfrac{1}{3}\) khi \(x=1\)

`(2x^2+3)^2-6`

Ta có:`x^2>=0AAx`

`=>2x^2>=0AAx`

`=>2x^2+3>=3AAx`

`=>(2x^2+3)^2>=9AAx`

`=>(2x^2+3)^2-6>=3AAx`

Dấu "=" xảy ra `<=>x=0`

Vậy `max_((2x^2+3)^2-6)=3` khi `x=0`