1.
$P= a^2 (c-b) + b^2(b-c)+c^2(1-c)$
$=(c-b)(a^2-b^2)+c^2(1-c)$
$ 0 \leq a \leq b \leq c \leq 1$
$ \rightarrow c-b \geq 0; \ a^2-b^2 \leq 0$$ \rightarrow (c-b)(a^2-b^2) \leq 0$
$ \rightarrow P \leq c^2(1-c)=4.\frac{c}{2}.\frac{c}{2}.(1-c)$
BĐT Cauchy 3 số $ (x+y+z)^3 \geq 27xyz \ \ (x,y,z \geq 0)$:
$ P \leq 4 \frac{(\frac{c}{2}+\frac{c}{2}+1-c)^3}{27}=\frac{4}{27}$
Vậy $ max P = \frac{4}{27} \leftrightarrow \left \{ {{\frac{c}{2}=1-c} \atop {\left[ \begin{array}{l}a=b\\b=c\end{array} \right. }} \right.$ $\leftrightarrow\left \{ {{c=\frac{2}{3}} \atop {\left[ \begin{array}{l}a=b\\b=c\end{array} \right. }} \right.$
$\leftrightarrow \left[ \begin{array}{l}b=c=\frac{2}{3};0\leq a \leq1 \\c=\frac{2}{3};0\leq a=b \leq1\end{array} \right. $
2.
$2(x^2+2x+3)=5\sqrt{x^3+3x^2+3x+2}$
$\leftrightarrow 2[(x+1)^2+2]=5\sqrt{ (x+1)^3+1} \ \ (1)$
$ĐK: (x+1)^3+1 \geq 0 \leftrightarrow x+1 \geq -1 \leftrightarrow x \geq -2$
Đặt $ x+1=a \ (a \geq -1)$
$(1) \leftrightarrow 2(a^2+2)=5\sqrt{ a^3+1} $
$\rightarrow 4.(a^2+2)^2=25( a^3+1) $
$\leftrightarrow 4a^4-25a^3+16a^2-9=0$
$\leftrightarrow (x^2-5x-3)(4x^2-5x+3)=0 \leftrightarrow ...$
