Question

Pe multimea Z se defineste operatia algebrica " * ",astfel:x*y=2x+y-3
a)Sa se calculeaze:4*7,8*(-1),(-8)*3 si 3*(-8).
b)Sa se afle valorile x e Z pentru care x* (3x-1)=6.
c)Se se rezolve ecuatia:(x+1)*3=5*(x pătrat-8).​

Answer 1

Explicație pas cu pas:

a)4*7=2×4+7-3=8+7-3=12

8*(-1)=2×8+(-1)-3=16-1-3=12

(-8)*3=2×(-8)+3-3=-16

3*(-8)=2×3+(-8)-3=6-8-3=-5

b)x*(3x-1)=6=>2x +3x-1-3=6 =>

5x-4=6 =>5x=10 =>x=2

c)(x+1)*3=5*(x²-6) =>

2(x+1)+3-3=2×5+(x²-6)-3 =>

2x+2=10+x²-6-3=>

2x+2=x²+1 =>x²-2x+1=2 =>(x-1)²=2 =>

x1-1=V2 sau x2-1=-V2 => x1=V2-1 sau

x2=1-V2

Answer 2

Răspuns:

Explicație pas cu pas:

x*y=2x+y-3

a) 4*7=2·4+7-3=8+7-3=12

8*(-1)=2·8+(-1)-3=16-1-3=12

(-8)*3=2·(-8)+3-3=-16

3*(-8)=2·3+(-8)-3=6-8-3=-5

b) x*(3x-1)=2·x+(3x-1)-3=2x+3x-1-3=5x-4. Deci 5x-4=6, ⇒5x=6+4, ⇒5x=10, ⇒x=10:5, ⇒x=2∈Z

c) (x+1)*3=2·(x+1)+3-3=2x+2

5*(x²-8)=2·5+(x²-8)-3=10+x²-8-3=x²+1

Deci 2x+2=x²+1, ⇒x²-2x-2+1=0, ⇒ x²-2x-1=0, ecuatie de gradul 2

Δ=(-2)²-4·1·(-1)=4+4=8>0, √8=√(4·2)=2√2

x1=(2-1√2)/2=2·(1-√2)/2=1-√2, atunci x2=1+√2

Deci x∈{1±√2).