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

Ta có

x 3   +   3 x 2   +   3 x   +   1   =   0   ⇔   ( x   +   1 ) 3   =   0

ó x + 1 = 0 ó x = -1

Vậy x = -1

Đáp án cần chọn là: A

`P(x)=\(4x^2+x^3-2x+3-x-x^3+3x-2x^2\)

`= (x^3-x^3)+(4x^2-2x^2)+(-2x-x+3x)+3`

`= 2x^2+3`

`Q(x)=`\(3x^2-3x+2-x^3+2x-x^2\)

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

`= -x^3+2x^2-x+2`

`P(x)-Q(x)-R(x)=0`

`-> P(X)-Q(x)=R(x)`

`-> R(x)=P(x)-Q(x)`

`-> R(x)=(2x^2+3)-(-x^3+2x^2-x+2)`

`-> R(x)=2x^2+3+x^3-2x^2+x-2`

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

`= x^3+x+1`

`@`\(\text{dn inactive.}\)

a: P(x)-Q(x)-R(x)=0

=>R(x)=P(x)-Q(x)

=2x^2+3+x^3-2x^2+x-2

=x^3+x+1

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

$\Leftrightarrow -2(-x^2+3x-4)=0$

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

Bài 1.

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

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

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

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

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))

c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)

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

      \(\Rightarrow x=1\)

b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)

     \(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)

     \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)

cdbdb

a.

\(\Leftrightarrow\left(x-1\right)^3=10^3\)

\(\Leftrightarrow x-1=10\)

\(\Rightarrow x=11\)

b.

\(\Leftrightarrow x^2-4x+4=25\)

\(\Leftrightarrow\left(x-2\right)^2=5^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

a) x³-3x²+3x-1=1000

⇒(x-1)³=1000

⇒x-1=10

⇒x=11

b) x²-4x-21=0

⇒ x²-7x+3x-21=0

⇒x(x-7)+3(x-7)=0

⇒(x+3)(x-7)=0

⇒ hoặc x+3 = 0 ⇒ x=-3

hoặc x-7=0⇒x=7

vậy x={-3; 7}

a) Có: \(x^3-3x^2.1+3.1^2x-1^3=1000\)

          \(\Rightarrow\left(x-3\right)^3=1000\)

          \(\Rightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

b) Có: \(x^2-4x+4-25=0\)

     \(\Rightarrow\left(x-2\right)^2=25\)

     \(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)