[tex] \frac{ x}{8 - x} = \frac{1}{ \sqrt{3} } < = > \\ x \sqrt{3} = 8 - x \\ x \sqrt{3} + x = 8 < = > \\ x( \sqrt{3} + 1) = 8 | \div 8 = > \\ x = ^{ \sqrt{3} - 1 )} \frac{8}{ \sqrt{3} + 1} < = > \\ x = 4( \sqrt{3} - 1) = > x = 4 \sqrt{3} - 4[/tex]
Răspuns: [tex]\bf x = 4\sqrt{3}-4[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf \dfrac{x}{8-x}=\dfrac{1}{\sqrt{3}}[/tex]
[tex]\bf \dfrac{x}{8-x}=\dfrac{1\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}[/tex]
[tex]\bf \dfrac{x}{8-x}=\dfrac{\sqrt{3}}{3}[/tex]
facem produsul mezilor egal cu produsul extremilor
[tex]\bf 3x = \sqrt{3} \cdot (8-x)[/tex]
[tex]\bf 3x = 8\sqrt{3} -\sqrt{3} x[/tex]
[tex]\bf 3x +\sqrt{3} x= 8\sqrt{3}[/tex]
[tex]\bf x \cdot(3 +\sqrt{3}) = 8\sqrt{3}[/tex]
[tex]\bf x = \dfrac{8\sqrt{3} }{3 +\sqrt{3}}[/tex]
[tex]\bf x = \dfrac{8\sqrt{3}\cdot (3 -\sqrt{3} ) }{(3 +\sqrt{3})\cdot (3 -\sqrt{3} )}[/tex]
[tex]\bf x = \dfrac{12\sqrt{3}-12}{3}[/tex]
[tex]\bf x = \dfrac{3 \cdot(4\sqrt{3}-4)}{3}[/tex]
[tex]\bf x = \dfrac{\not 3 \cdot(4\sqrt{3}-4)}{\not3}[/tex]
[tex]\boxed{\bf x = 4\sqrt{3}-4}[/tex]
sau
[tex]\boxed{\bf x = 4\cdot(\sqrt{3}-1)}[/tex]
==pav38==
