\(\Leftrightarrow x^3+3x^2+3x+1+27=0\\ \Leftrightarrow\left(x+1\right)^3+27=0\\ \Leftrightarrow\left(x+4\right)\left(x^2+2x+1-3x-3+9\right)=0\\ \Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}=0\left(vn\right)\end{matrix}\right.\Leftrightarrow x=-4\)

ĐKXĐ: \(x\ge\sqrt[3]{7}\)

\(4x^3-x^2+2x-32+\left(x^3-4\right)\left(\sqrt{x^3-7}-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^2+7x+16\right)+\dfrac{\left(x^3-4\right)\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^2+7x+16+\dfrac{\left(x^3-4\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}\right)=0\)

\(\Leftrightarrow x=2\) (ngoặc đằng sau luôn dương do \(x^3-4=x^3-7+3>0\))

2.

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

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

Đặt \(\left\{{}\begin{matrix}2x^3=a\\x+1=b\end{matrix}\right.\)

\(\Rightarrow a^3-b^3+a-b=0\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Rightarrow2x^3=x+1\Leftrightarrow\left(x-1\right)\left(2x^2+2x+1\right)=0\)

\(-2\left(2x+2\right)-3=28\)

\(-2\left(2x+2\right)=28+3\)

\(-2\left(2x+2\right)=30\)

\(2x+2=30:\left(-2\right)\)

\(2x+2=15\)

\(2x=15-2\)

\(2x=13\)

\(x=13:2\)

\(x=\dfrac{13}{2}\)

Vậy .................

\(\Leftrightarrow\dfrac{3x-1}{\left(6x-7\right)\left(3x+4\right)}-\dfrac{4x}{\left(8x-3\right)\left(3x+4\right)}=\dfrac{3}{\left(8x-3\right)\left(6x-7\right)}\)

=>(3x-1)(8x-3)-4x(6x-7)=3(3x+4)

=>24x^2-9x-8x+3-24x^2+28x=9x+12

=>11x+3=9x+12

=>2x=9

=>x=9/2

x³+3x²+3x+28=0

<=>x³+3x²+3x+1=-27

<=>(x+1)³=(-3)³

<=>x+1=-3

<=>x=-3-1

<=>x=-4

Vậy x=-4

Nếu theo Ác Mộng thì 3x² và 3x đi đâu rồi ???

Bạn Ác Mộng đúng mà. Bạn Ngọc không nhìn ra à?

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x+28=0\)

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

\(\Leftrightarrow x+1=-3\)

hay x=-4

\(=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-28=0\)

\(=>3x^2+26x=0=>x\left(3x+26\right)=0\)

\(=>\orbr{\begin{cases}x=0\\3x+26=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}}\)