Question

proszę o rozwiązanie

zad. 9 a,b (załącznik)

Answer 1

a)  zał. x≠ 0, x ≠ - 2

(6 - 3x) : (x + 2) / x = 1

(6 - 3x) * x / (x + 2) = 1

[ x * (6 - 3x) ] / (x + 2) = 1

x * (6 - 3x) =  (x + 2) * 1

6x - 3x² = x + 2

-3x² + 6x - x - 2 = 0

-3x² + 5x - 2 = 0

3x² - 5x + 2 = 0

Δ = 25 - 4 * 3 * 2 = 25 - 24 = 1

√Δ = 1

x1 = (5 - 1) / 6 = 4/6 = 2/3

x2 = (5 + 1)/6 = 6/6 = 1

 

odp. x = 2/3  lub  x = 1

 

b)

zał. x ≠ 5/2

(7x - 1) / (5 - 2x) : 3x = 1/2

(7x - 1) / (5 - 2x) * 1 / 3x = 1/2

(7x - 1) / [(5 - 2x) ]* 3x ] = 1/2

2 * (7x - 1) = 1 *  [(5 - 2x) ]* 3x ]

14x - 2 = 15x - 6x²

6x² + 14x - 15x - 2 = 0

6x² - x - 2 = 0

Δ = 1 - 4 * 6 *(-2) = 1 + 48 = 49

√Δ = 7

x1 = (1 - 7)/12 = -6/12 = -1/2

x2 = (1 + 7)/12 = 8/12 = 2/3

 

odp. x = -1/2  lub x = 2/3