Question

Пожалуйста, решите СРОЧНО!​

Answer 1

Ответ:

$a)\ \ \displaystyle \frac{1}{x^3+2}-\frac{1}{x^3+3}=\frac{1}{12} \ \ ,\ \ \ \ ODZ:\ x^3\ne -2\ ,\ x^2\ne -3\\\\\\\frac{12(x^3+3)-12(x^3+2)-(x^3+2)(x^2+3)}{12(x^3+2)(x^3+3)}=0\\\\\\\frac{36-24-(x^6+5x^3+6)}{12(x^3+2)(x^3+3)}=0\ \ ,\ \ \ x^6+5x^3+6=12\ \ ,\ \ x^6+5x^3-6=0\\\\\\(x^3)^2+5x^3-6=0\ \ ,\ \ (x^3)_1=-6\ \ ,\ \ (x^3)_2=1\ \ \ teorema\ Vieta\\\\x_1=-\sqrt[3]{6}\ \ ,\ \ x_2=\sqrt[3]1=1$

$b)\ \ \dfrac{21}{x^2-4x+10}-x^2+4x=6\ \ ,\ \ \ ODZ:\ x^2-4x+10\ne 0\\\\\\\dfrac{21}{(x^2-4x)+10}-(x^2-4x)=6\\\\\\y=x^2-4x\ \ \ \Rightarrow \ \ \ \dfrac{21}{y+10}-y-6=0\ \ ,\ \ \ \dfrac{21-y(y+10)-6(y+10)}{y+10}=0\\\\\\21-y^2-10y-6y-60=0\ \ ,\ \ y^2+16y+39=0\ \ ,\\\\D=16^2-4\cdot 39=100\ \ ,\ \ y_1=-13\ ,\ y_2=-3\\\\1)\ \ x^2-4x=-13\ \ ,\ \ x^2-4x+13=0\ \ ,\ \ D/4=2^2-13<0\ \ ,\ \ x\in \varnothing \\\\2)\ \ x^2-4x=-3\ \ ,\ \ x^2-4x+3=0\ \ ,\ \ x_1=1\ ,\ x_2=3\ (teorema\ Vieta)\\\\Otvet:\ x_1=1\ ,\ x_2=3\ .$

$c)\ \ \left\{\begin{array}{l}x^2+y^4=5\\xy^2=2\end{array}\right\ \ \left\{\begin{array}{l}x^2+(y^2)^2=5\\x\cdot y^2=2\end{array}\right\ \ \left\{\begin{array}{l}y^2=\dfrac{2}{x}\ ,\ x\ne 0\\\ x^2+\dfrac{4}{x^2}=5\end{array}\right\ \ \left\{\begin{array}{l}y^2=\dfrac{2}{x}\\\dfrac{x^4-5x^2+4}{x^2}=0\end{array}\right$

$\left\{\begin{array}{l}y^2=\dfrac{2}{x}\\(x^2)^2-5x^2+4=0\end{array}\right\ \ \left\{\begin{array}{l}y^2=\dfrac{2}{x}\\(x^2)_1=1\ ,\ (x^2)_2=4\end{array}\right\ \ \left\{\begin{array}{l}y^2=\dfrac{2}{x}\\x_{1,2}=\pm 1\ ,\ x_{3,4}=\pm 2\end{array}\right$

$x_1=-1\ ,\ y^2_1=-2\ \ \to \ \ \varnothing \\\\x_2=1\ \ ,\ \ y_2^2=2\ \ ,\ \ y_2=\pm \sqrt2\\\\x_3=-2\ \ ,\ \ y_3^2=-1\ \ \to \ \ \ \varnoting \\\\x_4=2\ \ ,\ \ y_4^2=1\ \ ,\ \ y_4=\pm 1\\\\Otvet:\ \ (\ 1\ ;-\sqrt2\ )\ ,\ (\ 1\ ;\, \sqrt2\ )\ ,\ (\ 2\ ;-1\ )\ ,\ (\ 2\ ;\ 1\ )\ .$