Question

За 100 ! 1)решите уравнение: 1/x+2-3/x+1=x/x^2+3x+2 2)x/x+3-x-4/3-x=1 умоляю вас! !

Answer 1

$\frac{1}{x+2}- \frac{3}{x+1}= \frac{x}{x^2+3x+2} \\ \frac{1}{x+2}- \frac{3}{x+1}= \frac{x}{(x+2)(x+1)} \\ \frac{x+1-3(x+2)-x}{(x+2)(x+1)}= 0 \\ \frac{x+1-3x-6-x}{(x+2)}= 0 \\ \frac{-3x-5}{(x+2)(x+1)}= 0 \\ -3x-5=0 \\ -3x=5 \\ x=-1 \frac{2}{3}$

$\frac{x}{x+3}- \frac{x-4}{3-x}=1 \\ \frac{x}{x+3}+\frac{x-4}{x-3}= \frac{x^2-9}{x^2-9} \\ \frac{x(x-3)+(x-4)(x+3)-x^2+9}{x^2-9}=0 \\ \frac{x^2-3x+x^2-x-12-x^2+9}{x^2-9}=0 \\ \frac{x^2-4x-3}{x^2-9}=0 \\ x^2-4x-3=0 \\ D=16+12=28 \\ \sqrt{D}= \sqrt{28}=2 \sqrt{7} \\ x_1= \frac{4+2 \sqrt{7} }{2} =2+ \sqrt{7} \\ x_2= \frac{4-2 \sqrt{7} }{2} =2- \sqrt{7}$