Question

5.67. a) (x + 3)^4 - 13 (x + 3)² + 36 = 0; 6) (2x - 1)^4 - (2x - 1)² - 12 = 0; с решением решить уравнение ​

Answer 1

Ответ:

a) $-6;\ -5;\ -1;\ 0$

6) $-\frac{1}{2};\ 1,5$

Объяснение:

a)

$(x + 3)^4 - 13 (x + 3)^2 + 36 = 0$

подставляем

$(x+3)^2=t,\ t\geq 0\\\\t^2-13t+36=0\\\\D=(-13)^2-4\cdot1\cdot36=169-144=25\\\\\sqrt{D}=\sqrt{25}=5\\\\t_1=\frac{13-5}{2\cdot1}=\frac{8}{2}=4\\\\t_2=\frac{13+5}{2\cdot1}=\frac{18}{2}=9\\\\(x+3)^2=4\\\\x+3=-2\ \ \ \ \ \ \ \ \ x+3=2\\\\x=-2-3\ \ \ \ \ \ \ \ \ x=2-3\\\\x=-5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-1\\\\\\(x+3)^2=9\\\\x+3=-3\ \ \ \ \ \ \ \ \ \ x+3=3\\\\x=-3-3\ \ \ \ \ \ \ \ \ \ x=2-3\\\\x=-6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-1$

6)

$(2x - 1)^4 - (2x - 1)^2 - 12 = 0$

мы подставляем

$(2x-1)^2=t,\ \ t\geq 0\\\\t^2-t-12=0\\\\D=(-1)^2-4\cdot1\cdot(-12)=1+48=49\\\\\sqrt{D}=\sqrt{49}=7\\\\t_1=\frac{1-7}{2\cdot1}=\frac{-6}{2}=-3 < 0\\\\t_2=\frac{1+7}{2\cdot1}=\frac{8}{2}=4\\\\(2x-1)^2=4\\\\2x-1=-2\ \ \ \ \ \ \ \ \ \ \ \ 2x-1=2\\\\2x=-2+1\ \ \ \ \ \ \ \ \ \ \ \ 2x=2+1\\\\2x=-1\ \ \ |:2\ \ \ \ \ \ \ \ \ \ 2x=3\ \ \ |:2\\\\x=-\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=1,5$

Answer 2

$\displaystyle\bf\\1)\\\\(x+3)^{4} -13(x+3)^{2} +36=0\\\\(x+3)^{2} =m \ , \ \ m \geq 0\\\\m^{2} -13m+36=0\\\\D=(-13)^{2} -4\cdot 36=169-144=25=5^{2} \\\\\\m_{1} =\frac{13-5}{2} =\frac{8}{2} =4\\\\\\m_{2} =\frac{13+5}{2} =\frac{18}{2} =9\\\\1)\\\\(x+3)^{2} =4\\\\\\\left[\begin{array}{ccc}x+3=-2\\x+3=2\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =-5\\x_{2} =-1\end{array}\right\\\\2)\\\\(x+3)^{2} =9\\\\\\\left[\begin{array}{ccc}x+3=-3\\x+3=3\end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}x_{3} =-6\\x_{4} =0\end{array}\right\\\\\\Otvet \ : \ -5 \ ; \ -1 \ ; \ -6 \ ; \ 0\\\\2)\\\\(2x-1)^{4} -(2x-1)^{2} -12=0\\\\(2x-1)^{2} =m \ , \ \ m \geq 0\\\\m^{2} -m-12=0$

$\displaystyle\bf\\D=(-1)^{2} -4\cdot (-12)=1+48=49=7^{2} \\\\\\m_{1} =\frac{1+7}{2} =\frac{8}{2} =4\\\\\\m_{2} =\frac{1-7}{2} =\frac{-6}{2} =-3 < 0 \ - \ ne \ podxodit\\\\(2x-1)^{2} =4\\\\\\\left[\begin{array}{ccc}2x-1=-2\\2x-1=2\end{array}\right\\\\\\\left[\begin{array}{ccc}2x=-1\\2x=3\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =-0,5\\x_{2} =1,5\end{array}\right\\\\\\Otvet \ : \ -0,5 \ ; \ 1,5$