Question

Tìm x, biết :

a) \(x^3-\dfrac{1}{4}x=0\)

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

c) \(x^2\left(x-3\right)+12-4x=0\)

Answer 1

Bài giải:

a) x³ – $\frac{1}{4}$x = 0 => x(x² –${\left(\frac{1}{2}\right)}^2$) = 0

=>x(x - $\frac{1}{2}$)(x + $\frac{1}{2}$) = 0

Hoặc x = 0

Hoặc x - $\frac{1}{2}$ = 0 => x = $\frac{1}{2}$

Hoặc x + $\frac{1}{2}$ = 0 => x = -$\frac{1}{2}$

Vậy x = 0; x = -$\frac{1}{2}$; x = $\frac{1}{2}$.

b) (2x – 1)² – (x + 3)² = 0

[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0

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

(x - 4)(3x + 2) = 0

Hoặc x - 4 = 0 => x = 4

Hoặc 3x + 2 = 0 => 3x = 2 => x = -$\frac{2}{3}$

Vậy x = 4; x = -$\frac{2}{3}$.

c) x²(x – 3) + 12 – 4x = 0

x²(x – 3) - 4(x -3)= 0

(x - 3)(x²- 2²) = 0

(x - 3)(x - 2)(x + 2) = 0

Hoặc x - 3 = 0 => x = 3

Hoặc x - 2 =0 => x = 2

Hoặc x + 2 = 0 => x = -2

Vậy x = 3; x = 2; x = -2.

Answer 2

a ) \(x^3-\dfrac{1}{4}x=0\)

\(\Leftrightarrow\) \(x\left(x^2-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

Hoặc x = 0

Hoặc \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Hoặc \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)

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

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

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

Hoặc \(x-4=0\Rightarrow x=4\)

Hoặc \(3x+2=0\Rightarrow3x=-2\Rightarrow x=-\dfrac{2}{3}\)

c) \(x^2 (x-3) + 12 - 4x = 0\)

\(\Leftrightarrow x^2\left(x-3\right)-\left(4x-12\right)=0\)

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

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

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

Hoặc \((x - 3) = 0\) \(\Rightarrow\) x = 3

Hoặc \(x - 2 = 0\) \(\Rightarrow\) x = 2

Hoặc \(x + 2 = 0 ​\) \(\Rightarrow\) x = \(- 2\)