a)g(x)=0=>11x3+5x2+4x+10=0

=>(10x3+10)+(x3+x2)+(4x2+4x)=0

=>10(x3+1)+x2(x+1)+4x(x+1)=0

=>10(x+1)(x2−x+1)+x2(x+1)+4x(x+1)=0

=>(x+1)[(10(x2−x+1)+x2+4x]=0

=>(x+1)(11x2−6x+10)=0

=>(x+1)[(9x2−2.3x+1)+2x2+9]=0

=>(x+1)[(3x−1)2+2x2+9]=0

=>x+1=0

=>x=-1

Vậy x=-1

Bài 1.

a.\(\left(x-8\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b.\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)

\(\Leftrightarrow4x-3-x-5=30-3x\)

\(\Leftrightarrow4x-x+3x=30+5+3\)

\(\Leftrightarrow6x=38\)

\(\Leftrightarrow x=\dfrac{19}{3}\)

Bài 1:

a. $(x-8)(x^3+8)=0$

$\Rightarrow x-8=0$ hoặc $x^3+8=0$

$\Rightarrow x=8$ hoặc $x^3=-8=(-2)^3$

$\Rightarrow x=8$ hoặc $x=-2$

b.

$(4x-3)-(x+5)=3(10-x)$

$4x-3-x-5=30-3x$

$3x-8=30-3x$

$6x=38$
$x=\frac{19}{3}$

Bài 2:

$f(x)=(x-1)(x+2)=0$

$\Leftrightarrow x-1=0$ hoặc $x+2=0$

$\Leftrightarrow x=1$ hoặc $x=-2$

Vậy $g(x)$ cũng có nghiệm $x=1$ và $x=-2$

Tức là:

$g(1)=g(-2)=0$

$\Rightarrow 1+a+b+2=-8+4a-2b+2=0$

$\Rightarrow a=0; b=-3$