a) \(x^3+9x^2+27x+19=0\)
\(\Rightarrow x^3+x^2+8x^2+8x+19x+19=0\)
\(\Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\)
Mà \(x^2+8x+19=x^2+2.x.4+16+3=\left(x+4\right)^2+3\)
Vì \(\left(x+4\right)^2\ge0\) với mọi x
\(3>0\)
\(\Rightarrow\left(x+4\right)^2+3>0\) với mọi x
=> ( x + 4 )2 + 3 vô nghiệm
=> x + 1 = 0
=> x = -1
Vậy x = -1
b) \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)
\(\Rightarrow\left(2x\right)^3+3.\left(2x\right)^2+3.2x+1+x\left(x^2-2^2\right)-9x\left(x^2-4x+4\right)+57=0\)
\(\Rightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)
\(\Rightarrow48x^2-34x+58=0\)
\(\Rightarrow2\left(24x^2-17x+29\right)=0\)
\(\Rightarrow24x^2-17x+29=0\)
... Tới đây mình bí luôn rồi, sorry 
Câu a : \(x^3+9x^2+27x+19=0\)
\(\Leftrightarrow\left(x^3+9x^2+27x+27\right)-8=0\)
\(\Leftrightarrow\left(x+3\right)^3-2^3=0\)
\(\Leftrightarrow\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+2^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Leftrightarrow x+1=0\) ( Vì : \(x^2+8x+19>0\))
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
Câu b : \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)
\(\Leftrightarrow48x^2-34x+58=0\)
\(\Rightarrow PTVN\)
Vậy ko có giá trị của x
a)Ta có: x^3 +9x^2 +27x + 19=0
x^3 + 9x^2 + 27x + 27 -8=0
(x+3)^3 -2^3=0
(x+3-2)(x^2 + 6x+9 +4x+12 +4)=0
(x+1)(x^2 +10x+ 25)=0
=> x+1=0 hoặc x^2 +10x+25=0
=> x=-1 hoặc (x+5)^2 =0
=> x=-1 hoặc x+5=0
=> x=-1 hoặc x=-5
