Question

Rezolvati in Q ecuatiile:

a) 1/2 * (x-1/2) + 1/3 * (x-1/3) = 1/6 * (x-1/6)
b) radical3 * x+1=x+radical3
c) x-2/3=4-x/2
d) x-radical5/radical5 = x-radical2/radical2
e) 3y+7/5 - 0.1 + 3-2y/2=0,3 *(4y-5)

Answer 1

a)

$\dfrac{1}{2}\Big(x-\dfrac{1}{2}\Big) + \dfrac{1}{3}\Big(x-\dfrac{1}{3}\Big) = \dfrac{1}{6}\Big(x-\dfrac{1}{6}\Big) \ \ \bigg| \cdot6$

$3\Big(x-\dfrac{1}{2}\Big) + 2\Big(x-\dfrac{1}{3}\Big) = x-\dfrac{1}{6}$

$3x-\dfrac{3}{2} + 2x-\dfrac{2}{3} = x-\dfrac{1}{6} \ \ \bigg| \cdot6$

$18x - 9 + 12x - 4 = 6x - 1$

$30x - 6x = 13 - 1 \iff 24x = 12$

$x = \dfrac{12}{24} \implies x = \dfrac{1}{2}$

b)

$\sqrt{3}x+1 = x+\sqrt{3}$

$\sqrt{3}x-x = \sqrt{3}-1$

$x(\sqrt{3} - 1) = \sqrt{3}-1 \ \ \Big|:(\sqrt{3}-1)$

$\implies x = 1$

c)

$\dfrac{x-2}{3} = \dfrac{4-x}{2}$

$2(x-2) = 3(4-x)$

$2x-4 = 12-3x \iff 2x+3x=12+4$

$5x=16 \implies x = \dfrac{16}{5}$

d)

$\dfrac{x-\sqrt{5} }{\sqrt{5}} = \dfrac{x-\sqrt{2} }{\sqrt{2}}$

$(x-\sqrt{5} ) \cdot \sqrt{2} = (x-\sqrt{2}) \cdot \sqrt{5}$

$x\sqrt{2} -\sqrt{10} = x\sqrt{5} - \sqrt{10}$

$x\sqrt{2} - x\sqrt{5} = \sqrt{10} - \sqrt{10}$

$x(\sqrt{2} - \sqrt{5}) = 0 \implies x = 0$