Question

1) (x-1)³+(2x+3)³=27x³+8

2) 3x³+2x²+2x+3=0

Giúp mk vs!!!!!!!

Answer 1

\(3x^3+2x^2+2x+3=0\)

\(\Leftrightarrow3\left(x^3+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow3\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2-x+3\right)=0\)

Mà \(3x^2-x+3=3\left[\left(x-\frac{1}{6}\right)^2+\frac{35}{36}\right]>0\forall x\)

Do đó: \(x+1=0\Leftrightarrow x=-1\)

Tập nghiệm: \(S=\left\{-1\right\}\)

Answer 2

\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow\left[\left(x-1\right)+\left(2x+3\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(6x^2-15x-9\right)=0\)(Chuyển vế)

\(\Leftrightarrow3\left(3x+2\right)\left(2x^2-5x-3\right)=0\)

\(\Leftrightarrow3\left(3x+2\right)\left(x-3\right)\left(2x+1\right)=0\)

Tập nghiệm: \(S=\left\{-\frac{2}{3};3;-\frac{1}{2}\right\}\)