Question

Tìm x, biết :

a) \(2-25x^2=0\)

b) \(x^2-x+\dfrac{1}{4}=0\)

Answer 1

Bài giải:

a) 2 – 25x² = 0 => (√2)² – (5x)² = 0

=> (√2 – 5x)( √2 + 5x) = 0

Hoặc √2 – 5x = 0 => 5x = √2 => x = $\frac{\sqrt{2}}{5}$

Hoặc √2 + 5x = 0 => 5x = -√2 => x = - $\frac{\sqrt{2}}{5}$

b) x² - x + $\frac{1}{4}$ = 0 => x² – 2 . x . $\frac{1}{2}$ + ($\frac{1}{2}$)² = 0

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

Answer 2

a) 2-25x²=0

<=>-25x²=-2

<=>25x²=2

<=>x²=\(\dfrac{2}{25}\)

<=>x=\(\sqrt{\dfrac{2}{25}}\)

b)x²-x +\(\dfrac{1}{4}\) =0

<=>(x - \(\dfrac{1}{2}\))² = 0

<=> x-\(\dfrac{1}{2}\) =0

<=>x=\(\dfrac{1}{2}\)