Question

. Va roooggg, repde6daca sr poate ​

Answer 1

Hei! :)

$a)\ (x-3)(2x-\sqrt{3})=0\\=> cazul\ 1: x-3=0\ sau\ =>\ 2x-\sqrt{3}=0\\=> x-3=0 => x=3\ sau =>\ 2x-\sqrt{3}=0 => x=\frac{\sqrt{3} }{2} \\Dar\ x\ apartine\ Q => x=\frac{\sqrt{3} }{2}$

$b) |x-1|=5\\cazul\ 1: x-1=5 => x=6\\cazul\ 2: x-1=-5=> x=-4 \\Dar\ x\ apartine\ N=> x=6$

$c) \sqrt{3} *x-3=3x-3\sqrt{3} \\\sqrt{3} *x-3x=3-3\sqrt{3}\\x(\sqrt{3} -3)=3-3\sqrt{3} \\x=\frac{3-3\sqrt{3}}{\sqrt{3} -3}$

Answer 2

Răspuns:

Explicație pas cu pas:

a)

(x - 3)(2x - √3) = 0 ⇒ x - 3 = 0 sau 2x - √3 = 0

⇒ x = 3 sau x = √3/2

solutia rationala este x = 3

b)

Ix - 1I = 5 ⇒ x - 1 = 5 sau x - 1 = -5

⇒ x = 6 sau x = -4

solutia naturala este x = 6

c)

√3*x - 3 = 3x - 3√3

3x - √3*x = 3√3 - 3

x*(3 - √3) = 3(√3 - 1)

x = 3(√3 - 1)/(3 - √3) = 3(√3 - 1)/√3(√3 - 1) = 3/√3 = √3