Question

Ayuden!
1) 7x² = 7
2) 2x² - 3x + 1 = 0
3) x² = 15x - 2x²
4) x² - 6x + 7 = 0 ​

Answer 1

Respuesta:

7x² = 7

$\frac{7x^2}{7}=\frac{7}{7}$

x^2=1

$x=\sqrt{1},\:x=-\sqrt{1}$

$x=1,\:x=-1$

2x² - 3x + 1 = 0

2x^2-3x+1=0

$x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:2\cdot \:1}}{2\cdot \:2}$

$x_{1,\:2}=\frac{-\left(-3\right)\pm \:1}{2\cdot \:2}$

$x=1,\:x=\frac{1}{2}$

x² = 15x - 2x²

intercambias lados :v

15x-2x²2=x2²

$15x-2x^2=x^2$

simplificas

$-3x^2+15x=0$

$x_{1,\:2}=\frac{-15\pm \sqrt{15^2-4\left(-3\right)\cdot \:0}}{2\left(-3\right)}$

$x_{1,\:2}=\frac{-15\pm \:15}{2\left(-3\right)}$

$x=0,\:x=5$

x² - 6x + 7 = 0 ​

$x^2-6x+7=0\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:7}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-\left(-6\right)\pm \:2\sqrt{2}}{2\cdot \:1}$

$x_1=\frac{-\left(-6\right)+2\sqrt{2}}{2\cdot \:1},\:x_2=\frac{-\left(-6\right)-2\sqrt{2}}{2\cdot \:1}$

$x=3+\sqrt{2},\:x=3-\sqrt{2}$

Explicación paso a paso:

espero te ayude