\(\left(x+1\right)^2\left(x+2\right)+\left(x-1\right)^2\left(x-2\right)=12\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x+2\right)+\left(x^2-2x+1\right)\left(x-2\right)=12\)\(\Leftrightarrow x^3+2x^2+2x^2+4x+x+2+x^3-2x^2-2x^2+4x+x-2-12=0\)\(\Leftrightarrow2x^3+10x-12=0\)
\(\Leftrightarrow2x^3-2x+12x-12=0\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+6\right)=0\)
Ta có :
\(x^2+x+6=x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\)\(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy phương trình có tập nghiệm \(S=\left\{1\right\}\)
(x+1)^2(x+2) + (x-1)^2(x-2) = 12
<=> (x^2 + 2x + 1)(x + 2) + (x^2 - 2x + 1)(x - 2) - 12 = 0
<=> x^3 + 2x^2 + 2x^2 + 4x + x + 2 + x^3 - 2x^2 - 2x^2 + 4x + x - 2 - 12 = 0
<=> 2x^3 + 10x - 12 = 0
<=> 2x^3 - 2x + 12x - 12 = 0
<=> 2x(x^2 - 1) + 12(x - 1) = 0
<=> 2(x - 1)[x(x + 1) + 6] = 0
<=> 2(x - 1)(x^2 + x + 6) = 0
Vì x^2 + x + 6 = x^2 + 2.x.1/2 + 1/4 + 23/4 = (x + 1/2)^2 + 23/4 > 0 với mọi giá trị của x
nên 2(x - 1)(x^2 + x + 6) = 0 <=> x - 1 = 0 hay x = 1
