Question

Помогите решить!! Более подробно нужно!! 100б

Answer 1

Ответ:

Объяснение:

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Answer 2

$5)\ \ \left\{\begin{array}{l}x(x+3y)=18\\y\, (3y+x)=6\end{array}\right\ \ \left\{\begin{array}{l}x+3y=\dfrac{18}{x}\\3y+x=\dfrac{6}{y}\end{array}\right\ \ \left\{\begin{array}{l}x+3y=\dfrac{18}{x}\\\dfrac{18}{x}=\dfrac{6}{y}\end{array}\right\\\\\\\left\{\begin{array}{l}x+3y=\dfrac{18}{x}\\18y=6x\end{array}\right\ \ \left\{\begin{array}{l}x+x=\dfrac{18}{x}\\3y=x\end{array}\right\ \ \left\{\begin{array}{ccc}2x^2=18\\x=3y\end{array}\right\ \ \left\{\begin{array}{l}x^2=9\\x=3y\end{array}\right$

$\left\{\begin{array}{ccc}x_1=-3\ ,\ x_2=3\\y_1=-1\ ,\ x_2=1\end{array}\right\ \ \ \Rightarrow \ \ \ n(x^2+y^2)=2\cdot (9+1)=2\cdot 10=20\\\\\\Otvet:\ \ \underline {\ n(x^2+y^2)=20\ }\ .$

$6)\ \ log_2\dfrac{x-2}{x+2}+log_{\frac{1}{2}}\dfrac{2x-1}{6x+7}=0\\\\\\ODZ:\ \left\{\begin{array}{l}\dfrac{x-2}{x+2}>0\\\dfrac{\, 2x-1}{6x+7}>0\end{array}\right\ \ \left\{\begin{array}{ccc}x\in (-\infty ;=2)\cup (2;+\infty )\\x\in (-\infty ;-\dfrac{7}{6})\cup (\dfrac{1}{2};+\infty )\end{array}\right\ \ \Rightarrow \\\\\\x\in (-\infty ;-2)\cup (2;+\infty )\\\\log_2\dfrac{x-2}{x+2}-log_2\dfrac{2x-1}{6x+7}=0\ \ ,\ \ log_2\dfrac{x-2}{x+2}=log_2\dfrac{2x-1}{6x+7}$

$\dfrac{x-2}{x+2}=\dfrac{2x-1}{6x+7}\ \ \ ,\ \ \ \ \dfrac{x-2}{x+2}-\dfrac{2x-1}{6x+7}=0\ \ ,\\\\\\\dfrac{(x-2)(6x+7)-(2x-1)(x+2)}{(x+2)(6x+7)}=0\ \ ,\ \ \ \dfrac{6x^2-5x-14-(2x^2+3x-2)}{(x+2)(6x+7)}=0\ ,\\\\\\4x^2-8x-12=0\ \ ,\ \ x^2-2x-3=0\ \ ,\ \ x_1=-1\ ,\ x_2=3\ \ (teorema\ Vieta)\\\\x_1=-1\notin ODZ\\\\Otvet:\ \ \underline {\ x=3\ }\ .$

$4x^2-2x-12=0\ \ ,\ \ 2x^2-x-6=0\ \ ,\ \ D=49\ ,\\\\x_1=-\dfrac{3}{2}\ \ ,\ \ x_2=2\notin ODZ$