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

a) Thay đa thức này bằng 0, ta được: 

f(x) = x^3 - x^2 + x - 1 = 0

=> f(x) = x . x² - x . x + x - 1 = 0

=> f(x) = x. (x² - x + x) = 0 + 1 = 1

=> f(x) = x . x² = 1

=> x = 1   và    x² = 1

=> x = 1

Vậy nghiệm của đa thức là x = 1

Đa thức \(h\left(x\right)=x^3+3x^2+3x+1.\)có nghiệm 

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x^2+1=0\left(ktm\right)\end{cases}\Rightarrow x=-\frac{1}{3}}\)

Vậy   .........

Ta có: \(h\left(x\right)=0\Leftrightarrow x^3+3x^2+3x+1=0\) 

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

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

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

                                 \(\Leftrightarrow\left[\left(x^2+x\right)+\left(x+1\right)\right].\left(x+1\right)=0\)

                                  \(\Leftrightarrow\left[x.\left(x+1\right)+\left(x+1\right)\right].\left(x+1\right)=0\)

                                 \(\Leftrightarrow\left(x+1\right).\left(x+1\right).\left(x+1\right)=0\)

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

                                    \(\Leftrightarrow x+1=0\)

                                    \(\Leftrightarrow x=-1\)

Vậy...

sorry nhìn nhầm 

Đa thức h(x) có nghiệm 

<=> x³+3x²+3x+1=0

<=> x³+2x²+x²+2x+x+1=0

<=> x².(x+1)+2x.(x+1)+(x+1)=0

<=>(x+1).(x²+2x+1)=0

<=> (x+1)³=0

<=> x+1=0

<=> x=-1

Vậy............

a: a(x)=x^3+3x^2+5x-18

b(x)=-x^3-3x^2+2x-2

b: m(x)=a(x)+b(x)

=x^3+3x^2+5x-18-x^3-3x^2+2x-2

=7x-20

c: m(x)=0

=>7x-20=0

=>x=20/7

\(3x^3+4x^2+2x+1=0\)

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

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

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

Ta có:\(3x^2+x+1=3\left(x^2+x.\frac{1}{3}+\frac{1}{3}\right)\)

\(=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{1}{3}\right)\)

\(=3\left[\left(x+\frac{1}{6}\right)^2+\frac{11}{36}\right]\ge3.\frac{11}{36}=\frac{11}{12}>0\forall x\)

Do đó x + 1 = 0 tức là x = -1

\(3x^3+3x^2+x^2+x+x+1=0\)

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

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

+)\(3x^2+x+1=0\Leftrightarrow3.\left(x^2+x+\frac{1}{3}\right)=0\Leftrightarrow3.\left(x+\frac{1}{6}\right)^2+\frac{11}{12}=0\left(loai\right)\)

+) x+1=0 <=> x=-1

t sao cl gì v??

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:

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

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

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