Question

Tìm x:
a) x⁴-25x³=0
b) (x-5)²-(3x-2)²=0
c) x³-4x²-9x+36=0
d) (-x³+3x²-4x) : (\(-\dfrac{1}{2}\)x)=0

Answer 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.

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

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

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

Answer 2

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.

Answer 3

Answer 4

a) \(x^4-25x^3=0\Leftrightarrow x^3\left(x-25\right)=0\)

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

b) \(\left(x-5\right)^2-\left(3x-2\right)^2=0\Leftrightarrow\left(x-5-3x+2\right)\left(x-5+3x-2\right)=0\Leftrightarrow-\left(2x+3\right)\left(4x-7\right)=0\)

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

c) \(x^3-4x^2-9x+36=0\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-3\right)=0\)

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

 

Answer 5

a: Ta có: \(x^4-25x^3=0\)

\(\Leftrightarrow x^3\cdot\left(x-25\right)=0\)

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

b: Ta có: \(\left(x-5\right)^2-\left(3x-2\right)^2=0\)

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

\(\Leftrightarrow\left(-2x+3\right)\left(4x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{7}{4}\end{matrix}\right.\)

Answer 6

c: Ta có: \(x^3-4x^2-9x+36=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)\left(x-3\right)=0\)

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

d: Ta có: \(\dfrac{-x^3+3x^2-4x}{-\dfrac{1}{2}x}=0\)

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

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

\(\Leftrightarrow x=0\left(loại\right)\)