Question

помогите пожалуйста ​даю 50 баллов

Answer 1

1)

$x^4-4x^2+4=0\\y=x^2 = > \\y^2-4y+4 = 0\\(y-2)^2 = 0\\y = 2\\x^2 = 2\\x1 = \sqrt{2} , x2=-\sqrt{2}$

Ответ: $x1 = \sqrt{2} , x2=-\sqrt{2}$

2)

$|3x^2+14x|-5=0\\y = 3x^2+14x\\ = > \\\left \{ {{y=5} \atop {-y=5}} \right. < = > \left \{ {{3x^2+14x=5} \atop {-3x^2-14x=5}} \right.\\\\a) 3x^2+14x=5\\3x^2+14x-5=0\\x1 = \frac{-14+\sqrt{196+60}}{6} = \frac{-14+16}{6} = \frac{1}{3}\\\\x2 = \frac{-14-\sqrt{196+60}}{6} = \frac{-14-16}{6} = -5\\\\b) -3x^2-14x=5\\-3x^2-14x-5=0\\x3 = \frac{-14+\sqrt{196-60}}{6} = \frac{-14+\sqrt{136}}{6} = \frac{-7+\sqrt{34}}{3}\\x4 = \frac{-14-\sqrt{196-60}}{6} = \frac{-14-\sqrt{136}}{6} = \frac{-7-\sqrt{34}}{3}$

Ответ: $x1 = \frac{1}{3}, x2=-5, x3=\frac{-7+\sqrt{34}}{3}, x4=\frac{-7-\sqrt{34}}{3}$

3)

$5x^2-8|x|+4=0 < = > \left \{ {{5x^2-8x+4=0} \atop {5x^2+8x+4=0}} \right. \\a) 5x^2-8x+4=0\\x1 = \frac{8+\sqrt{64-60}}{10} = \frac{8+2}{10} =\frac{10}{10} = 1\\x2 = \frac{8-\sqrt{64-60}}{10} = \frac{8-2}{10} =\frac{6}{10} = \frac{3}{5}\\\\b) 5x^2+8x+4=0\\x3 = \frac{-8+\sqrt{64-60}}{10} = \frac{-8+2}{10} =\frac{-6}{10} = -\frac{3}{5}\\x4 = \frac{-8-\sqrt{64-60}}{10} = \frac{-8-2}{10} =\frac{-10}{10} = -1$

Ответ: $x1=1, x2=\frac{3}{5},x3=-\frac{3}{5},x4=-1$