Đáp án D

x 3 − 3 x 2 + 4 x − 12 = 0 ⇔ x 2 x − 3 + 4 x − 3 = 0 ⇔ x − 3 x 2 + 4 = 0 ⇔ x = 2 i x = − 2 i x = 3 .

Vậy  z 1 − z 2 = 0.

Nếu đề bài là phân tích đa thức thành nhân tử thì làm như sau:

           \(x^3-3x^2-4x+12\)

\(=x ^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

\(=\left(x-3\right)\left(x^2+2x-2x-4\right)\)

\(=\left(x-3\right)\left[x\left(x+2\right)-2\left(x+2\right)\right]\)

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

Nếu đề là tìm nghiệm thì làm đến đó ta làm tiếp:

\(\Leftrightarrow\)\(x-3=0\)\(\Leftrightarrow\)\(x=3\)

hoặc   \(x+2=0\)\(\Leftrightarrow\)\(x=-2\)

hoặc  \(x-2=0\)\(\Leftrightarrow\)\(x=2\)

x 3 - 3 x 2 - 4 x + 12 = x 3 - 3 x 2 - 4 x - 12 = x 2 x - 3 - 4 x - 3 = x - 3 x 2 - 4 = x - 3 x + 2 x - 2

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: \(x^2-4x+4-25=0\)

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

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

Để phân thức

bằng 0

⇒

Chọn đáp án C.

Để phân thức  bằng 0 ⇔

Chọn đáp án C.

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.

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=x\left(x-y\right)+2\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=x^2\left(x-3\right)-4\left(x-3\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

a) \(xy^2-25x=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\)

b) \(x\left(x-y\right)+2x-2y=x\left(x-y\right)+\left(2x-2y\right)=x\left(x-y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+2\right)\)

c) \(x^3-3x^2-4x+12=\left(x^3-3x^2\right)-\left(4x-12\right)=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=x\left(x-y\right)+2\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

\(x\left(y^2-25\right)=y\left(y-5\right)\left(y+5\right)\)

\(x\left(x-y\right)2\left(x-y\right)=\left(x-y\right)2x\)

\(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)