Question

doy corona.
con procedimiento
ayuda xfavor :'vvv
​

Answer 1

Respuesta:

Explicación paso a paso:

1) 2x² + 1 = 9

  2x² +1 - 9 = 0

  2x² - 8 = 0

  2(x² - 4) = 0

     x² - 2² = 0

 (x + 2)(x - 2) = 0

 x + 2 = 0  =>  $x_{1}$ = - 2

 x  - 2 = 0  =>  $x_{2}$ =  2   resp (c)

2)  2x² = 72

      x² = 72/2

      x² - 36 = 0

  x² - 6² = 0

 (x + 6)(x - 6) = 0

 x + 6 = 0  =>  $x_{1}$ = - 6

 x  - 6 = 0  =>  $x_{2}$ =  6  resp (e)

3)  4x² = 4

    x² = 1

  x² - 1 = 0

 x² - 1² = 0

 (x + 1)(x - 1) = 0

 x + 1 = 0  =>  $x_{1}$ = - 1

 x  - 1 = 0  =>  $x_{2}$ =  1    resp (a)

4)  $\frac{x^{2} }{2}=2$

 x² = 2*2 = 4

 x² - 2² = 0

 (x + 2)(x - 2) = 0

 x + 2 = 0  =>  $x_{1}$ = - 2

 x  - 2 = 0  =>  $x_{2}$ =  2   resp (b)

5)  $x^{2} - \frac{1}{4} = 0$

$x^{2} - \frac{1}{2^{2} } = 0$

$x^{2} - (\frac{1}{2})^{2} = 0$

$(x +\frac{1}{2})(x -\frac{1}{2}) = 0$

x + $\frac{1}{2}$ = 0  =>  $x_{1}$ = - $\frac{1}{2}$

 x  - $\frac{1}{2}$ = 0  =>  $x_{2}$ =  $\frac{1}{2}$   resp (c)

6) x² - $\frac{4}{9}$ = 0

$x^{2} - \frac{2^{2} }{3^{2} } = 0$

$x^{2} - (\frac{2}{3 })^{2} = 0$

$(x +\frac{2}{3})(x -\frac{2}{3}) = 0$

x + $\frac{2}{3}$ = 0  =>  $x_{1}$ = - $\frac{2}{3}$

 x  - $\frac{2}{3}$ = 0  =>  $x_{2}$ =  $\frac{2}{3}$   resp (d)

7)  25x² - 36 = 0

(5x)² - 6² = 0

(5x + 6)(5x - 6) = 0  

5x + 6 = 0  =>  $x_{1}$ = - $\frac{6}{5}$

 5x  - 6 = 0  =>  $x_{2}$ =  $\frac{6}{5}$   resp (e)

8) (x + 2)(x - 6) = 0

 x + 2 = 0  =>  $x_{1}$ = - 2

 x  - 6 = 0  =>  $x_{2}$ =  6  resp (c)

9) 5x² - 7x - 6 = 0 ±

ax² + bx + c = 0   =>    $x = \frac{-b\± \sqrt{b^{2}-4ac } }{2a}$

a = 5

b = -7

c = -6

reemplaza

$x = \frac{-(-7)\± \sqrt{(-7)^{2}-4(5)(-6) } }{2(5)}$

$x = \frac{7\± \sqrt{49+120} }{10}=\frac{7\± \sqrt{169} }{10}=\frac{7\± 13 }{10}$

$x_{1}=\frac{7\+ 13 }{10} = \frac{20}{10}= 2$

$x_{2}=\frac{7\ - 13 }{10} = \frac{-6}{10}=- \frac{3}{5}$   resp (d)