Matemática, perguntado por hasteinanonykpb7ijh, 11 meses atrás

Resolva as equacões do 2 grau

(A) 2 (y+6) = y(y-2)

(B) x (X + 3) = 40

(C) m²/4-1=5m/8-5/4

(D) 2y-2/5+1/10-y(y+1)/5=-3/10

Soluções para a tarefa

Respondido por Pitágoras1618
3

A) 2(y+6)=y(y-2)

2y+12=y²-2y

2y+2y-y²+12=0

4y-y²+12=0


a= -1

b= 4

c= 12


Y= -b±√b²-4ac/2a


Y= -4±√4²-4.(-1).12/2.(-1)

Y= -4±√16+48/-2

Y= -4±√64/-2

Y= -4±8/-2


Y'= -4+8/-2= -2

Y"= -4-8/-2= 6


B) x(x+ 3) = 40

x²+3x=40

x²+3x-40=0


a= 1

b= 3

c= -40


Δ= b²-4ac

Δ= 3²-4.1.(-40)

Δ=9+160

Δ=169


X= -b±√Δ/2.a

X= -3±√169/2.1

X= -3±13/2


X'= -3+13/2= 5

X"= -3-13/2= -8


C)  \frac{m^{2}}{4-1}  = \frac{\frac{5m}{8-5}}{4}


 \frac{4m^2}{3} = \frac{5m}{3}


4m²=5m

4m²-5m=0


4m[m-(5/4)]=0


4m=0

m=0


m-5/4=0

m= 5/4


M'=0

M"=5/4


D) 2y-\frac{2}{5}+\frac{1}{10} -\frac{y^2+y}{5} = \frac{-3}{10}


M.M.C(5,10)= 2×5=10


 \frac{20y}{10} -\frac{4}{10} +\frac{1}{10} -\frac{2.(y^2+y)}{10} =\frac{-3}{10}


20y-4+1-2y²-2y= -3

18y-2y²= -3+4-1

18y-2y²=0


2y(y+9)=0


2y=0

y=0


y+9=0

y= -9


y'=0

y"= -9


Bons estudos!



hasteinanonykpb7ijh: Mano muito obrigado mesmo!
Pitágoras1618: De nada!
Pitágoras1618: * A resposta da D) é y'=0 e y"=9.
hasteinanonykpb7ijh: Opa vlw <3
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