Question

REZOLVA EXERCITIILE
1)
c)x supra 5 - 3-x supra 10=x+3 supra 2 ,x apartine Q
d) |2x+4|=8
2) Rezolvati in multimea numerelor reale ecuatiile
a)√3(x-2)+2√3(x+1)=6√3
b)2-√3+2x(2-√3)=4(x-√3)+√27

Answer 1

Explicație pas cu pas:

$\mathbf{c) \frac{x}{5} - \frac{3 - x}{10} = \frac{x + 3}{2} }$

$\mathbf{2x - (3 - x) = 5(x + 3)}$

$\mathbf{2x - 3 + x = 5x + 15}$

$\mathbf{3x - 3 = 5x + 15}$

$\mathbf{3x - 5x = 15 + 3}$

$\mathbf{ - 2x = 18}$

$\mathbf{x = 18 \div ( - 2) = > x = - 9}$

$\mathbf{d) |2x + 4| = 8 }$

$\mathbf{2x + 4 = 8 = > 2x = 4 = > x 1= 2}$

$\mathbf{2x + 4 = - 8 = > 2x = - 12 = > x = - 6}$

$\mathbf{a) \sqrt{3}(x - 2) + 2 \sqrt{3} (x + 1) = 6 \sqrt{3} }$

$\mathbf{ \sqrt{3}x - 2 \sqrt{3} + 2 \sqrt{3} x + 2 \sqrt{3} = 6 \sqrt{3} }$

$\mathbf{3 \sqrt{3} x = 6 \sqrt{3} }$

$\mathbf{x = 6 \sqrt{3} \div (3 \sqrt{3}) = > x = 2 }$

$\mathbf{b)2 - \sqrt{3} + 2x(2 - \sqrt{3} ) = 4(x - \sqrt{3}) + \sqrt{27} }$

$\mathbf{2 - \sqrt{3} + 4x - 2 \sqrt{3} x = 4x - 4 \sqrt{3} + 3 \sqrt{3} }$

$\mathbf{2 - \sqrt{3} - 2 \sqrt{3} x = - 4 \sqrt{3} + 3 \sqrt{3} }$

$\mathbf{2 - \sqrt{3} - 2 \sqrt{3}x = - \sqrt{3} }$

$\mathbf{2 - 2 \sqrt{3}x = 0 }$

$\mathbf{x = \frac{1}{ \sqrt{3} } = > x = \frac{ \sqrt{3} }{3} }$

Bafta!