a: B\A=(-1;4]

\(C_R^B=R\text{\B}=(-\infty;-1]\cup\left(6;+\infty\right)\)

b: C=(-2;4]

D={0}

\(C\cap D=(-2;4]\)

Ta có: ab+bc+ca=abc

nên abc-ab-bc-ac=0

Ta có: a+b+c=1

nên a+b+c-1=0

Ta có: abc-ab-bc-ac+a+b+c-1=0

\(\Leftrightarrow\left(abc-ab\right)-\left(bc-b\right)-\left(ac-a\right)+\left(c-1\right)=0\)

\(\Leftrightarrow ab\left(b-1\right)-b\left(c-1\right)-a\left(c-1\right)+\left(c-1\right)=0\)

\(\Leftrightarrow b\left(c-1\right)\left(a-1\right)-\left(c-1\right)\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)

\(A=(-\infty;-3]\cup[-4;+\infty)\)

B=(-vô cực,2) giao (5;+vô cực)

1: A hợp B=(-vô cực,2) giao [-4;+vô cực]=R

A\B=[-4;5]

2: (B\A) giao N=(-3;2) giao N=[2;+vô cực)

Xét a⁴ - 2a³ \(\ge8a-16\)

<=> a⁴ -2a³ -8a +16\(\ge0\)

<=> (a⁴ - 2a³) - 8 (a-2) \(\ge0\)

<=> \(a^3\left(a-2\right)-8\left(a-2\right)\ge0\)

<=> \(\left(a-2\right)\left(a^3-8\right)\ge0\)

<=> \(\left(a-2\right)^2\left(a^2+2a+4\right)\ge0\) (luôn đúng)

Tương tự => \(\left\{{}\begin{matrix}b^4-2b^3\ge8b-16\\c^4-2c^3\ge8c-16\end{matrix}\right.\)

<=> \(a^4+b^4+c^4-2\left(a^3+b^3+c^3\right)\ge8\left(a+b+c\right)-48=0\)

<=> \(a^4+b^4+c^4\ge2\left(a^3+b^3+c^3\right)\)

Dấu "=" <=> a=b=c=2