Question

(х^2+х) ^2-6(х^2+х) +8=0​

Answer 1

$\bf (x^2+x) ^2-6(x^2+x) +8=0$

$t {}^{2} - 6t + 8 = 0$

$\fbox{ t = 2 \\ t = 4}$

$\fbox{x {}^{2} + x = 2 \\ x {}^{2} + x = 4}$

$\fbox{x = - 2 \\ x = 1 \\ x = \dfrac{ - 1 + \sqrt{17} }{2} \\ x = \dfrac{ - 1 - \sqrt{17} }{2} }$

Ответ:

$\fbox{x_{1} = - 2 \\ x_{2} = 1 \\ x_{3} = \dfrac{ - 1 + \sqrt{17} }{2} \\ x_{4} = \dfrac{ - 1 - \sqrt{17} }{2} }$

Answer 2

Ответ:

Пошаговое объяснение:

x²+x=t

t²-6t+8=0

t1=4; t2=2

1)x²+x=4⇒x²+x-4=0⇒D=1+16=17⇒x1=(-1+√17)/2; x2=(-1-√17)/2

2)x²+x=2⇒x²+x-2=0⇒x3=1; x4=-2