Question

1)x²+2|x|-3=0
2)x²-5|x|+1=0​

Answer 1

Ответ:

1) $\boxed{\left[ \begin{gathered} x_{1,2} = б3 \\ x_{3,4} = б1\end {gathered}}$

2) $\boxed{\left[ \begin{gathered} x_{1,2} = б\sqrt{ \dfrac{23 + \sqrt{525} }{2}} \\ x_{3,4} = б \sqrt{ \dfrac{23 - \sqrt{525} }{2}}\end {gathered}}$

Объяснение:

1) $x^{2} + 2|x| - 3 = 0$

$2|x| = 3 - x^{2}$

$(2|x|)^{2} = (3 - x^{2})^{2}$

$4x^{2} = 9 - 6x^{2} + x^{4}$

$x^{4} - 10x^{2} + 9 = 0$

Замена: $x^{2} = t; t\geq 0$

$t^{2} - 10t + 9 = 0$

$D = 100 - 4 \cdot 1 \cdot 9 = 100 - 36 = 64 = 8^{2}$

$t_{1} = \dfrac{10 + 8}{2} = \dfrac{18}{2} = 9$

$t_{2} = \dfrac{10 - 8}{2} = \dfrac{2}{2} = 1$

$\left[ \begin{gathered} x^{2} = t_{1} \\ x^{2} = t_{2} \end {gathered}$  $\left[ \begin{gathered} x^{2} = 9 \\ x^{2} = 1 \end {gathered}$ $\left[ \begin{gathered} \sqrt{x^{2} } = \sqrt{9} \\ \sqrt{x^{2} } = \sqrt{1}\end {gathered}$ $\left[ \begin{gathered} |x| = 3 \\ |x| = 1\end {gathered}$ $\left[ \begin{gathered} x_{1,2} = б3 \\ x_{3,4} = б1\end {gathered}$

2) $x^{2} - 5|x| + 1 = 0$

$5|x| = x^{2} + 1$

$(5|x|)^{2} = (x^{2} + 1)^{2}$

$25x^{2} = x^{4} + 2x^{2} + 1$

$x^{4} - 23x^{2} + 1 = 0$

Замена: $x^{2} = t; t\geq 0$

$t^{2} - 23t + 1 = 0$

$D = 529 - 4 \cdot 1 \cdot 1 = 529 - 4 = 525$

$t_{1} = \dfrac{23 + \sqrt{525} }{2}$

$t_{2} = \dfrac{23 - \sqrt{525} }{2}$

$\left[ \begin{gathered} x^{2} = t_{1} \\ x^{2} = t_{2} \end {gathered}$  $\left[ \begin{gathered} x^{2} = \dfrac{23 + \sqrt{525} }{2} \\ x^{2} = \dfrac{23 - \sqrt{525} }{2} \end {gathered}$ $\left[ \begin{gathered} \sqrt{x^{2} } = \sqrt{ \dfrac{23 + \sqrt{525} }{2}} \\ \sqrt{x^{2} } = \sqrt{ \dfrac{23 - \sqrt{525} }{2}}\end {gathered}$ $\left[ \begin{gathered} |x| = \sqrt{ \dfrac{23 + \sqrt{525} }{2}} \\ |x| = \sqrt{ \dfrac{23 - \sqrt{525} }{2}}\end {gathered}$ $\left[ \begin{gathered} x_{1,2} = б\sqrt{ \dfrac{23 + \sqrt{525} }{2}} \\ x_{3,4} = б \sqrt{ \dfrac{23 - \sqrt{525} }{2}}\end {gathered}$