Equivalent disequations:\(\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)

Put \(x^4-11x^2+10=t\)

We have:t(t+18)<0

TH1:\(\left\{{}\begin{matrix}t>0\\t+18< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t>0\\t< -18< 0\end{matrix}\right.\) (loại)

TH2:\(\left\{{}\begin{matrix}t< 0\\t+18>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t< 0\\t>-18\end{matrix}\right.\)\(\Rightarrow t\in\left\{-17,-16,...........,-1\right\}\)

It's easy here, you try it yourself......

C1:10 + 20 + 30 +40+50+60

=(10+60)+(20+50)+(30+40)

=70+70+70

=70.3=210

C2:10+20+30+40+50+60

=10.1+10.2+10.3+10.4+10.5+10.6

=10.(1+2+3+4+5+6)

=10.21=210

We have:\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{4}{a+b+c}\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=4\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)\(\Rightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\Leftrightarrow\dfrac{a^2+ab+ca}{b+c}+\dfrac{ba+b^2+cb}{c+a}+\dfrac{ac+cb+c^2}{a+b}=a+b+c\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{c^2}{a+b}+\dfrac{b^2}{a+c}+a+b+c=a+b+c\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)