Question

{ab=12 e a=b+1

Como resolver?

Answer 1

Boa noite

Substituindo a por b+1 temos : (b+1)*b=12⇒ b² +b =12 ⇒b²+b-12=0

Δ=1²-4*1*(-12)=1+48 = 49 ⇒√Δ=7

$b=\frac{-1\pm\sqrt{49}}{2*1}=\frac{-1\pm7}{2} \\ \\ \\ \left \{ {{b'=\frac{-1*7}{2}=3} \atop {b''=\frac{-1-7}{2}=-4}} \right.$

a*3=12⇒a=12/3⇒a=4⇒(4,3)

a*(-4)=12⇒a=12/(-4)⇒a=-3⇒(-3,-4)

Resposta : S={ (4,3),(-3,-4)}

Answer 2

ab=12

a=b+1

A=12/b

12/b=b+1

b(b+1) =12

b² +b -12=0

Δ=1² -4( 1(-12)

Δ=1+48

Δ= 49

X= -b+-√Δ/2a

-1+-√49/2

-1+-7/2

X,=-1+7/2

X,=6/2

X,=3

X,,= -1-7/2

X,,= -8/2

X,,= -4 descartar esta(negativa)

a=b+1

a= 3+1

A=4 e B=3 = prova= ab=12 = (4x3)=12;ok