Question

determinatii solutiile ecuatiilor >
B) √5(x+4)+2√5(x-2)=√15
C) x(2√2-√3)+2x(√3-√2)=6+√3
D) x√2+2x(√8-3)=20-6x
E) 2x(2+√3)-(√3-1)(x+2)=2x+6
F) 2(x+√2-√5)+3=x(√5-√2)
La raspunsuri da - B) √3/3
C) 1+2√3
E) 3+√3 /3

Answer 1

$\displaystyle a). \sqrt{5} (x+4)+2 \sqrt{5} (x-2)= \sqrt{15} \\ \sqrt{5} x+4 \sqrt{5} +2 \sqrt{5} x-4 \sqrt{5} = \sqrt{15} \\ \sqrt{5} x+2 \sqrt{5} x= \sqrt{15} -4 \sqrt{4} +4 \sqrt{5} \\ 3 \sqrt{5} x= \sqrt{15} \\ x= \frac{ \sqrt{15} }{3 \sqrt{5} } \\ x= \frac{ \sqrt{3} }{3}$
$\displaystyle b).x(2 \sqrt{2} - \sqrt{3} )+2x( \sqrt{3} - \sqrt{2} )=6+ \sqrt{3} \\ 2 \sqrt{2} x- \sqrt{3} x+2 \sqrt{3} x-2 \sqrt{2} x=6+ \sqrt{3} \\ \sqrt{3} x=6+ \sqrt{3} \\ x= \frac{6+ \sqrt{3} }{ \sqrt{3} } \Rightarrow x= \frac{6 \sqrt{3} +3}{3} \Rightarrow x= \frac{3(2 \sqrt{3} +1)}{3} \Rightarrow x=2 \sqrt{3} +1$
$\displaystyle c).x \sqrt{2} +2x( \sqrt{8} -3)=20-6x \\ x \sqrt{2} +2x(2 \sqrt{2} -3)=20-6x \\ x \sqrt{2} +4x \sqrt{2} -6x=20-6x \\ x \sqrt{2} +4x \sqrt{2} -6x+6x=20 \\ 5x \sqrt{2} =20 \\ x= \frac{20}{5 \sqrt{2} } \Rightarrow x= \frac{4}{ \sqrt{2} } \Rightarrow x= \frac{4 \sqrt{2} }{2} =2 \sqrt{2}$
$\displaystyle d).2x(2+ \sqrt{3} )-( \sqrt{3} -1)(x+2)=2x+6 \\ 4x+2 \sqrt{3} x-( \sqrt{3} x+2 \sqrt{3} -x-2)=2x+6 \\ 4x+2 \sqrt{3} x- \sqrt{3} x-2 \sqrt{3} +x+2=2x+6 \\ 4x+2 \sqrt{3} x- \sqrt{3} x+x-2x=6+2 \sqrt{3} -2 \\ 3x+ \sqrt{3} x=4+2 \sqrt{3} \\ x(3+ \sqrt{3} )=4+2 \sqrt{3}$
$\displaystyle x= \frac{4+2 \sqrt{3} }{3+ \sqrt{3} } \Rightarrow x= \frac{(3- \sqrt{3} )(4+2 \sqrt{3} )}{9-3} \Rightarrow x= \frac{12+6 \sqrt{3}-4 \sqrt{3} -6 }{6} \Leftrightarrow \\ \\ \Leftrightarrow x= \frac{6+2 \sqrt{3} }{6} \Rightarrow x= \frac{2(3 + \sqrt{3} )}{6} \Rightarrow x= \frac{3+ \sqrt{3} }{3}$
$\displaystyle e).2(x+ \sqrt{2} - \sqrt{5} )+3=x( \sqrt{5} - \sqrt{2} ) \\ 2x+2 \sqrt{2} -2 \sqrt{5} +3= \sqrt{5} x- \sqrt{2} x \\ 2x- \sqrt{5} x+ \sqrt{2} x=2 \sqrt{5} -2 \sqrt{2} -3 \\ x(2- \sqrt{5} +\sqrt{2} )=2 \sqrt{5} -2 \sqrt{2} -3 \\ x= \frac{2 \sqrt{5} -2 \sqrt{2} -3}{2- \sqrt{5} +\sqrt{2} }$