Question

Решите верхнее и нижнее уравнение. Сколько не пробовал - не получается. ответ на верхнее: -2 ответ на нижнее: 12​

Answer 1

$1)\ \ \sqrt{1-4x}+2=\sqrt{(2x+1)^2-8x}\\\\ODZ:\ \left\{\begin{array}{l}1-4x\geq 0\\(2x+1)^2-8x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\leq 0,25\\(2x-1)^2\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\leq 0,25\\x\in R\end{array}\right\ \ x\leq 0,25\\\\\\1-4x+4\cdot \sqrt{1-4x}+4=(2x+1)^2-8x\\\\5-4x+4\cdot \sqrt{1-4x}=4x^2-4x+1\ \ \ ,\ \ \ 4\sqrt{1-4x}=4x^2-4\ \ ,\\\\\sqrt{1-4x}=x^2-1\ \ ,\ \ x^2-1\geq 0\ ,\ \ x\in (-\infty ;-1\ ]\cup [\ 1\ ;+\infty )$

$\left\{\begin{array}{l}x\in (-\infty ;-1\ ]\cup [\ 1\ ;+\infty )\\x\leq 0,25\end{array}\right\ \ \ x\in (-\infty ;-1\ ]$

$1-4x=x^4-2x^2+1\\\\x^4-2x^2+4x=0\ \ ,\ \ x\, (x^3-2x+4)=0\ \ ,\\\\a)\ \ x_1=0\ \ ne \ podxodit$

$b)\ \ x^3-2x+4=0\ \ \Rightarrow \ \ x=-2\ ,\ tak\ kak\ \ (-2)^3-2\cdot (-2)+4=0\ ,\\\\x^3-2x+4=(x+2)(x^2-2x+2)\\\\x^2-2x+20\ ,\ tak\ kak\ D=4-8=-4$

$2)\ \ \sqrt{\dfrac{20+x}{x}}+\sqrt{\dfrac{20-x}{x}}=\sqrt{6}\\\\\\ODZ:\ \ \left\{\begin{array}{l}\dfrac{20+x}{x}\geq 0\\\dfrac{20-x}{x}\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;-20\ ]\cup (\ 0\ ;+\infty )\\x\in (\ 0\ ;\ 20\ ]\end{array}\right\ \ \ \ x\in (\ 0\ ;\ 20\ ]\\\\\\\dfrac{20+x}{x}+2\cdot \sqrt{\dfrac{20+x}{x}\cdot \dfrac{20-x}{x}}+\dfrac{20-x}{x}=6$

$2\cdot \sqrt{\dfrac{20+x}{x}\cdot \dfrac{20-x}{x}}+\dfrac{40}{x}=6\ \ \ ,\ \ \ 2\cdot \sqrt{\dfrac{400-x^2}{x^2}}=6-\dfrac{40}{x}\\\\\\\sqrt{\dfrac{400-x^2}{x^2}}=3-\dfrac{20}{x}\ \ ,$

$3-\dfrac{20}{x}\geq 0\ \ ,\ \ \dfrac{3x-20}{x}\geq 0\ \ ,\ \ x\in (-\infty ;0\ )\cup [\ 6\dfrac{2}{3} \ ;+\infty )\\\\x\in [\ 6\dfrac{2}{3} \ ;\ 20\ ]\\\\\\\dfrac{400-x^2}{x^2}=9-\dfrac{120}{x}+\dfrac{400}{x^2}\ \ \ \Big|\cdot x^2\\\\\\400-x^2=9x^2-120x+400\ \ \ ,\ \ \ 10x^2-120x=0\ \ ,\\\\10x\cdot (x-12)=0\ \ \Rightarrow \ \ \ x_1=0\notin ODZ\ \ ;\ \ \ \ x-12=0\ ,\ x_2=12\\\\Otvet:\ x=12\ .$