Rozwiąż równania. a) (x-5)²/5-3=(x-2√5)(x+2√5)/5+x- Zadanie 12: Matematyka wokół nas 2

Rozwiązanie

$a) \frac{( x - 5 ) ^{2}}{5} - 3 = \frac{( x - 2 5 ) ( x + 2 5 )}{5} + x$

Skorzystamy ze wzorów:

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

$(a - b) (a + b) = a^{2} - b^{2}$

$\frac{x ^{2} - 2 \cdot x \cdot 5 + 5 ^{2}}{5} - \frac{15}{5} = \frac{x ^{2} - ( 2 5 ) ^{2}}{5} + \frac{5 x}{5}$

$\frac{x ^{2} - 10 x + 25 - 15}{5} = \frac{x ^{2} - 4 \cdot 5 + 5 x}{5} ∣ \cdot 5$

$x^{2} - 10 x + 10 = x^{2} - 20 + 5 x$

$- 10 x - 5 x = - 20 - 10$

$- 15 x = - 30 ∣ : (- 15)$

$x = 2$

$b) \frac{1}{2} (x 2 - 1) (x 2 + 1) = (x + 2)^{2} + x$

Skorzystamy ze wzorów:

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

$(a - b) (a + b) = a^{2} - b^{2}$

$\frac{1}{2} ((x 2)^{2} - 1^{2}) = (x^{2} + 2 \cdot x \cdot 2 + 2^{2}) + x$

$\frac{1}{2} (x^{2} \cdot 2 - 1) = x^{2} + 4 x + 4 + x$

$\frac{1}{2} (2 x^{2} - 1) = x^{2} + 5 x + 4$

$x^{2} - \frac{1}{2} = x^{2} + 5 x + 4$

$- \frac{1}{2} - 4 = 5 x ∣ : 5$

$- 4 \frac{1}{2} : 5 = x$

$x = - \frac{9}{2} : 5$

$x = - \frac{9}{2} \cdot \frac{1}{5}$

$x = - \frac{9}{10}$

$c) \frac{( x - 3 ) ( x + 3 )}{2} + 3 (2 - x) - \frac{5}{6} x^{2} = \frac{( 4 - x ) ^{2} - 2 x ( x - 4 )}{3} + 2 \frac{1}{6}$

Skorzystamy ze wzorów:

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

$(a - b) (a + b) = a^{2} - b^{2}$

$\frac{x ^{2} - ( 3 ) ^{2}}{2} + 3 (2 - x) - \frac{5}{6} x^{2} = \frac{4 ^{2} - 2 \cdot 4 \cdot x + x ^{2} - 2 x ^{2} + 8 x}{3} + \frac{13}{6} ∣ \cdot 6$

$6^{3} \cdot \frac{x ^{2} - 3}{2 ^{1}} + 6 \cdot (6 - 3 x) - 6^{1} \cdot \frac{5}{6 ^{1}} x^{2} = 6^{2} \cdot \frac{16 - 8 x + x ^{2} - 2 x ^{2} + 8 x}{3 ^{1}} + 6^{1} \cdot \frac{13}{6 ^{1}}$