Question

Решите уравнение пожалуйста ​

Answer 1

Объяснение:

б)

$(y^2-3y-5)(y^2-3y+1)=-5.$

Пусть y²-3y-5=t       ⇒

$t*(t+6)=-5\\t^2+6t+5=0\\D=16\ \ \ \ \ \sqrt{D}=4.\\ t_1=y^2-3y-5=-5\\y^2-3y=0\\y*(y-3)=0\\y_1=0\\y-3=0\\y_2=3.\\t_2=y^2-3y-5=-1\\y^2-3y-4-0\\D=25\ \ \ \ \sqrt{D}=5\\ y_3=-1\ \ \ \ \ y_4=4.$

Ответ: y₁=0,  y₂=3,  y₃=-1,  y₄=4.

в)

$\frac{y}{y+2}+\frac{y+2}{2-y}=\frac{8}{y^2-4}\\ \frac{y}{y+2}-\frac{y+2}{y-2}=\frac{8}{(y+2)(y-2)}.$

ОДЗ: у+2≠0  у≠-2  у-2≠0  у≠2.

$y*(y-2)-(y+2)*(y+2)=1*8\\y^2-2y-(y+2)^2=8\\y^2-2y-y^2-4x-4=8\\-6y=12\ |:(-6)\\y=-2\notin.\ \ \ \ \ \Rightarrow$

Ответ: y ∈ ∅.

 

Answer 2

Решим первое уравнение методом замены .

$\displaystyle\bf\\(y^{2} -3y-5)(y^{2} -3y+1)=-5\\\\y^{2} -3y-5=m \ \ \ \Rightarrow \ \ \ y^{2} -3y+1=m+6\\\\m(m+6)=-5\\\\m^{2} +6m+5=0\\\\Teorema \ Vieta:\\\\m_{1} +m_{2} =-6\\\\m_{1} \cdot m_{2} =5\\\\m_{1} =-1\\\\m_{2} =-5\\\\1) \ y^{2} -3y-5=-1\\\\y^{2}-3y-4=0\\\\Teorema \ Vieta:\\\\y_{1} =4 \ \ ; \ \ y_{2} =-1\\\\\ 2) \ y^{2} -3y-5=-5\\\\y^{2}-3y=0\\\\y\cdot(y-3)=0\\\\y_{3} =0\\\\y-3=0 \ \ \ \Rightarrow \ \ \ y_{4} =3$

$\displaystyle\bf\\Otvet: \ 4 \ ; \ - 1 \ ; \ 0 \ ; \ 3\\\\\\2)\\\\\frac{y}{y+2} +\frac{y+2}{2-y} =\frac{8}{y^{2}-4 } \\\\\\\frac{y}{y+2} -\frac{y+2}{y-2} -\frac{8}{(y+2)(y-2)} =0\\\\\\\frac{y\cdot(y-2)-(y+2)\cdot(y+2)-8}{(y+2)(y-2)} =0\\\\\\\frac{y^{2}-2y-y^{2}-2y-2y-4-8 }{(y+2)(y-2)} =0\\\\\\\frac{-6y-12}{(y+2)(y-2)} =0\\\\\\\left\{\begin{array}{ccc}-6y-12=0\\y+2\neq 0\\y-2\neq 0\end{array}\right\\\\\\\left\{\begin{array}{ccc}-6y=12\\y\neq -2\\y\neq 2\end{array}\right$

$\displaystyle\bf\\\left\{\begin{array}{ccc}y=-2\\y\neq -2\\y\neq 2\end{array}\right\\\\\\Otvet: \ \oslash$   Корней нет