Question

ezolvati ecuatiile
h) (2+√3)x-2+√3=0;

Answer 1

Răspuns:

Explicație pas cu pas:

(2+√3)x-2+√3=0

(2+√3)x=2-√3

$x=\frac{2-\sqrt{3} }{2+\sqrt{3} }$

rationalizam, adica amplificam cu 2-√3

$x=\frac{(2-\sqrt{3})^2 }{2^2-\sqrt{3}^2 }=\frac{(2-\sqrt{3})^2 }{1 }=7-4\sqrt{3}$

Answer 2

Explicație pas cu pas:

$\bf\red\:(2 + \sqrt{3} )x - 2 + \sqrt{3} = 0 \\\bf\red\:(2 + \sqrt{3} )x = 2 - \sqrt{3} \: \: \: | \: \: \div (2 + \sqrt{3} )\: \: \: \\ \bf\red\:x = (2 - \sqrt{3} ) \div (2 + \sqrt{3} ) \\ \bf\red\:x = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \\ \\ \bf\red\: x = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \bf\red\:x = \frac{(2 - \sqrt{3} ) \times (2 - \sqrt{3} )}{4 - 3} \\ \\ \bf\red\:x = \frac{(2 - \sqrt{3} ) \times (2 - \sqrt{3} )}{1} = > x = (2 - \sqrt{3} ) \times (2 - \sqrt{3} ) = > x = (2 - \sqrt{3} ) {}^{2} \\ \bf\red\:x = 4 - 4 \sqrt{3} + 3 = > \\ \bf\red\: x = 7 - 4 \sqrt{3}$