Respostas

2013-05-28T13:53:05-03:00

Primeiramente faça isso x²-9x=0 é a mesma coisa que x(x-9)=0

agora você vai achar a raiz da função que é x-9=0 -> x=9

vamos verificar

9²-9.9=0

81-81=0

 

16y²=9 -> y^{2}=\frac{9}{16}\\ y=\sqrt\frac{9}{16}\\y=\sqrt\frac{3^{2}}{4^{2}}\\y=\frac{3}{4}

 

vamos testar 1616.\frac{3^{2}}{4^{2}}=9\\\\\frac{16.9}{16}=9

 

c)x²-49=0 então x²=49 ->x=\sqrt49\\\\x=7

vamos testar 7²-49=0

49-49=0

 

d)m²-m=0 (-1)²+(-1)=0; 1-1=0

 

e)3x²-60=0 -> 3x²=60 -> x²=20 -> x=raiz de 20

 

f)6y²-2y=0 -> (isole o y) y(6y-2)=0 ach a raiz 6y-2=0, 6y=2, y=2/6, y=1/3.

vamos testar 

6y^{2}-2y=0\\\\\frac{6.1^{2}}{3^{2}}-\frac{2.1}{3}=0\\\\\frac{6}{9}-\frac{2}{3}=0\\\\simplificando\ 6\ e\ 9\ por\ 3\ temos\ \frac{2}{3}\\logo\ \frac{2}{3}-\frac{2}{3}=0

  • Usuário do Brainly
2013-05-28T21:12:11-03:00

a) \text{x}^2-9\text{x}=0

 

\text{x}\cdot(\text{x}-9)=0

 

\text{x}'=0

 

\text{x}"=9

 

 

b) 16\text{y}^2=9

 

\text{y}^2=\dfrac{9}{16}

 

\text{y}=\pm\sqrt{\dfrac{9}{16}}=\pm\dfrac{3}{4}

 

 

c) \text{x}^2-49=0

 

\text{x}^2=49

 

\text{x}=\pm\sqrt{49}=\pm7

 

 

 

d) \text{m}^2+\text{m}=0

 

\text{m}\cdot(\text{m}+1)=0

 

\text{m}'=0

 

\text{m}"=-1

 

 

e) 3\text{x}^2-60=0

 

3\text{x}^2=60

 

\text{x}^2=20

 

\text{x}=\pm\sqrt{20}=\pm2\sqrt{5}

 

 

 

f) 6\text{y}^2-2\text{y}=0

 

2\text{y}\cdot(3\text{y}-1)=0

 

\text{y}'=0

 

\text{y}"=\dfrac{1}{3}