Question

Решите уравнение (x+3)⁴+2(x+3)²-8=0
(x²-3)(7x+5)=0

Answer 1

$\displaystyle\bf\\1) \ \ (x+3)^{4} +2(x+3)^{2} -8=0\\\\(x+3)^{2} =m \ , \ m\geq 0 \ \ \Rightarrow \ \ (x+3)^{4} =m^{2} \\\\m^{2} +2m-8=0\\\\D=2^{2} -4\cdot (-8)=4+32=36=6^{2} \\\\m_{1} =\frac{-2+6}{2} =2\\\\m_{2} =\frac{-2-6}{2} =-4<0-neyd\\\\(x+3)^{2} =2\\\\x+3=\pm \sqrt{2} \\\\x_{1} =-3-\sqrt{2} \\\\x_{2} =-3+\sqrt{2}$

$\displaystyle\bf\\2) \ \ (x^{2} -3)(7x+5)=0\\\\\\\left[\begin{array}{ccc}x^{2} -3=0\\7x+5=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x^{2} =3\\7x=-5\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =-\sqrt{3} \\x_{2} =\sqrt{3} \\x_{3}=-\dfrac{5}{7} \end{array}\right$