Question

1. Stabiliti daca ecuatiile urmatoare admit solutia -2: a. x+3=1; b. x-4=-2; c. -2x+1=5; d. 2x+1=-x-5 2.Rezolvati urmatoarele ecuatii: a. x-3=-1,x={-1,0,1,2,3}; b.2x+7=3,x={-1,0,1}; c. -3x+1=-5, x={-2,-1,0,1,2,3}; d. 2-2x=4,x={-3,-2,-1} Va rog! Dau coroana si dau si 60 de puncte

Answer 1

Răspuns:

Explicație pas cu pas:

1. Inlocuim pe x cu -2:

a)-2+3 = 1, ok, avem solutie

b) -2-4 = -6 ≠ -2, nu este

c) -2*(-2) + 1 = 4+1 = 5, verifica, deci x = -2 este solutie a ecuatiei date

d) 2*(-2) + 1 = -4 + 1 = -3

-(-2) - 5 = 2 - 5 = -3, deci membrul 1 = membrul 2 pt x = -2, deci se verifia solutia x = -2.

2.

a) x-3=-1, x = 3-1 = 2

b) 2x+7=3, 2x = -4, x = -2

c) -3x+1=-5, 3x = 5+1, 3x = 6, x = 2

d) 2-2x=4, 2x = 2-4, 2x = -2, x = -1.

Answer 2

$\boxed{1} \ \bold{a)} \ \underline{-2}+3=1 , \ \Rightarrow \ se \ admite; \\ \bold{b)} \ \underline{-2}-4 \neq-2, \ \Rightarrow \ nu \ se \ admite; \\ \bold{c)} \ -2 \cdot \underline{(-2)}+1=4+1=5, \ \Rightarrow \ se \ admite; \\ \bold{d)} \ 2 \cdot \underline{(-2)}+1=-\underline{(-2)}-5 \\ -4+1=2-5 \ \Leftrightarrow \ -3=-3, \ \Rightarrow \ se \ admite.$

$\boxed{2} \ \bold{a)} \ x-3=-1 \ \Leftrightarrow \ x=-1+3=2; \\ \Rightarrow x \in\{-1,0,1,2,3\}; \\ \bold{b)} \ 2x+7=3 \ \Leftrightarrow \ 2x=-4 \ \Leftrightarrow \ x=-2; \\ \Rightarrow x \notin \{-1,0,1\};\\ \bold{c)} \ -3x+1=-5 \Leftrightarrow \ -3x=-6 \ \Leftrightarrow \ x=2; \\ \Rightarrow x \in \{-2,-1,-,1,2,3\}; \\ \bold{d)} \ 2-2x=4 \ \Leftrightarrow \ 2x=-2 \ \Leftrightarrow \ x=-1; \\ \Rightarrow x \in \{-3,-2,-1\}.$