Ta có (a+2)³-(a+6)(a²+12)+64=a³+6a²+12a+8-a³-12a-6a²-72+64=0(đpcm)

\(\left(a+2^3\right)-\left(a+6\right).\left(a^2+12\right)+64=0\)

\(\Leftrightarrow\left(a+8\right)-\left(a^3+6a^2+12a+72\right)=-64\)

\(\Leftrightarrow\left(a^3+6a^2+12a+72\right)-\left(a+8\right)=64\)

\(\Leftrightarrow a^3+6a^2+11a+64=64\)

\(\Leftrightarrow a^3+6a^2+11a^2=0\)

\(\Leftrightarrow a.\left(a^2+6a+11\right)=0\)

\(\Leftrightarrow a.\left[\left(a^2+2.a.3+9\right)+2\right]=0\)

\(\Leftrightarrow a.\left[\left(a+3\right)^2+2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\\left(a+3\right)^2+2=0\left(\text{Vô lí}\right)\end{matrix}\right.\)

\(\Rightarrow a=0\)

\(\Rightarrow\) Đpcm.

\(\Leftrightarrow\left(a+8\right)-\left(a^3+6a^2+12a+72\right)=-64\Leftrightarrow\left(a^3+6a^2+12a+72\right)-\left(a+8\right)=64\)

\(\Leftrightarrow a^3+6a^2+11a+64=64\Leftrightarrow a^3+6a^2+11a=0\Leftrightarrow a\left(a^2+6a+11\right)=0\)

\(\Leftrightarrow a\left[\left(a^2+2.a.3+9\right)+2\right]=0\Leftrightarrow a\left[\left(a+3\right)^2+2\right]=0\Leftrightarrow\orbr{\begin{cases}a=0\\\left(a+3\right)^2+2=0\left(V\text{ô}l\text{í}\right)\end{cases}\Rightarrow a=0}\)

hình như sai đề hay sao ý, có nghiệm mà =)))))

1/ x^3+6x^2+12x+8=0

(x+2)^3=0

x+2=0

x=-2

Vậy x=-2

a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

x²-2x+2=(x²-2x+1)+1=( x-1)²+1

Mà (x-1)²≥0 với mọi x

=> (x-1)²+1>0 với mọi x

=> x²-2x+2>0 với mọi x