Question

За 2 примера с решением даю 20 баллов.

Answer 1

$1. \ \frac{x+1}{x} - \frac{2+3x}{x^2+2x} = \frac{1-x}{x+2} \\ \frac{x+1}{x} - \frac{2+3x}{x(x+2)} -\frac{1-x}{x+2}=0 \\\\ \frac{(x+1)(x+2)}{x} - \frac{2+3x}{x(x+2)} - \frac{x(1-x)}{x+2}=0 \\ \frac{x^2+2x+x+2}{x(x+2)} - \frac{2+3x}{x(x+2)} - \frac{x-x^2}{x(x+2)}=0 \\ \frac{x^2+3x+2-(2+3x)-(x-x^2)}{x(x+2)} =0\\ x \neq 0, \ x \neq -2\\ x^2+3x+2-2-3x-x+x^2=0\\ 2x^2-x=0\\ x(2x-1)=0\\ x=0,\ no \ u \ nas \ D(y): \ x \neq 0\\ 2x-1=0\\ x=0,5$

$2. \ \frac{1-4x}{4x+1} = \frac{12}{1-16x^2} + \frac{1+4x}{4x-1} \\ \frac{1-4x}{4x+1} = \frac{12}{(1-4x)(1+4x)} - \frac{1+4x}{1-4x} \\ \frac{(1-4x)(1-4x)}{4x+1} = \frac{12}{(1-4x)(1+4x)} - \frac{(1+4x)(1+4x)}{1-4x} \\ \frac{1-4x-4x+16x^2}{(1-4x)(1+4x)} = \frac{12}{(1-4x)(1+4x)} - \frac{1+4x+4x+16x^2}{(1-4x)(1+4x)} \\ (1-4x)(1+4x) \neq 0, \ x \neq 0,25, \ x \neq -0,25\\ 1-8x+16x^2=12-(1+8x+16x^2)\\ 1-8x+16x^2=12-1-8x-16x^2\\ 16x^2+16x^2-8x+8x+1-11=0\\ 32x^2-10=0/:2\\ 16x^2-5=0$
$(4x- \sqrt{5} )(4x+\sqrt{5} )=0\\ 4x=\pm\sqrt{5} , \ x=\pm \frac{\sqrt{5} }{4}$