Răspuns :
1. daca 3√2x + 7 = 0 ; 3√2x = - 7 ; x = - 7 / 3√2 = - 7√2 / 3√2√2 =
= - 7√2 / 3·2 = - 7√2 / 6 ∈R
2)16 x-3(x+1)=5x
16x - 3x - 3 -5x = 0 ; 8x = 3 ; x = 3 / 8 ∈Q
3) 2,5(x-4)-6x=3-x
2,5x - 10 - 6x + x = 3
-2,5 x = 3 +10
-2,5 x = 13
x = 13 : ( -2,5 ) = - 5,2∈ Q
= - 7√2 / 3·2 = - 7√2 / 6 ∈R
2)16 x-3(x+1)=5x
16x - 3x - 3 -5x = 0 ; 8x = 3 ; x = 3 / 8 ∈Q
3) 2,5(x-4)-6x=3-x
2,5x - 10 - 6x + x = 3
-2,5 x = 3 +10
-2,5 x = 13
x = 13 : ( -2,5 ) = - 5,2∈ Q
[tex]\displaystyle a).3 \sqrt{2} x+7=0 \\ 3 \sqrt{2} x=0-7 \\ 3 \sqrt{2} x=-7 \\ x=- \frac{7}{3 \sqrt{2} } \\ x=- \frac{7 \sqrt{2} }{6} \in R[/tex]
[tex]\displaystyle b).16x-3(x+1)=5x \\ 16x-3x-3=5x \\ 16x-3x-5x=3 \\ 8x=3 \\ x= \frac{3}{8} \in Q[/tex]
[tex]\displaystyle c).2,5(x-4)-6x=3-x \\ 2,5x-10-6x=3-x \\ 2,5x-6x+x=3+10 \\ -2,5x=13 \\ x=- \frac{13}{2,5} \\ x=- 5,2 \in Q[/tex]
[tex]\displaystyle b).16x-3(x+1)=5x \\ 16x-3x-3=5x \\ 16x-3x-5x=3 \\ 8x=3 \\ x= \frac{3}{8} \in Q[/tex]
[tex]\displaystyle c).2,5(x-4)-6x=3-x \\ 2,5x-10-6x=3-x \\ 2,5x-6x+x=3+10 \\ -2,5x=13 \\ x=- \frac{13}{2,5} \\ x=- 5,2 \in Q[/tex]