\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3abc-3ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\left(đpcm\right)\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\left(đpcm\right)\)

\(\dfrac{2x-1}{2}+\dfrac{5-x}{6}=2-\dfrac{3\left(x+1\right)}{4}\)

\(\Leftrightarrow\dfrac{6\left(2x-1\right)}{12}+\dfrac{2\left(5-x\right)}{12}=\dfrac{24}{12}-\dfrac{9\left(x+1\right)}{12}\)

\(\Leftrightarrow6\left(2x-1\right)+2\left(5-x\right)=24-9\left(x+1\right)\)

\(\Leftrightarrow12x-6+10-2x=24-9x-9\)

\(\Leftrightarrow19x=11\)

`<=>x=11/19`

Vậy \(S=\left\{\dfrac{11}{19}\right\}\)

\(\dfrac{2x-1}{2}\) + \(\dfrac{5-x}{6}\) = 2 - \(\dfrac{3\left(x+1\right)}{4}\)

6(2x-1) + 2(5-x) = 24 - 9 (x+1)  (quy đồng mẫu số)

12x - 6 + 10 -2x = 24 - 9x - 9   (nhân phá ngoặc)

10x + 9x = 24 - 9  -10 + 6 ( chuyển vế đổi dấu)

19x = 11

x = 11/19

P = -x² +6x + 1 = - (x- 3)² + 10 ≤10 ⇔ P(max) = 10 ⇔ x =3