Question

Помогите решить, пожалуйста​

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\sqrt{\frac{2x}{x+1} }-\sqrt{\frac{2(x+1)}{x} } =1\\\\x\in(-\infty;-1)\cup (0;+\infty)\\\\\\\sqrt{2}\cdot \sqrt{\frac{x}{x+1} }-\sqrt{2}\cdot \sqrt{\frac{x+1}{x} } =1\\\\t=\sqrt{\frac{x}{x+1} }>0\\\\\sqrt{2}*t-\frac{\sqrt{2} }{t} =1\ \ \ \ |*t\\\\\sqrt{2}*t^2-t-\sqrt{2} =0\\\\D=b^2-4ac=1+4*\sqrt{2} *\sqrt{2} =9\\\\t_1=\frac{1-3}{2*\sqrt{2} } <0;ne\ udovl.\\\\t_2=\frac{1+3}{2*\sqrt{2}}=\frac{2}{\sqrt{2} }\\\\ \sqrt{\frac{x}{x+1} }=\frac{2}{\sqrt{2} }\\\\\frac{x}{x+1} =\frac{4}{2}=2$

$\displaystyle\\2x+2=x\\\\x=-2$

О т в е т : x = - 2

Answer 2

√(2x/(x + 1)) - √(2(x + 1)/x) = 1

одз

x/(x + 1) > 0

+++++++(-1) ------------------- (0) ++++++++

x ∈ (-∞,-1) U (0, +∞)

√(2x/(x + 1)) - √(2(x + 1)/x) = 1

√2*(√x/(x + 1)) - √((x + 1)/x)) = 1

√x/(x + 1) = y > 0

√2(y - 1/y) = 1

y² - 1 - y/√2 = 0

D = 1/2 + 4 = 9/2

y12 = (1/√2 +-√9/2)/2

y1 = (1/√2 - √9/2)/2 < 0 нет

y2 = (1/√2 + 3/√2)/2 = 4/√2 : 2 = 2/√2 = √2

√x/(x + 1) = y

√x/(x + 1) = √2

x/(x + 1) = 2

x = 2(x + 1)

x = 2x + 2

x = -2