Question

выполните действия.

Answer 1

$(\frac{x+1}{x^2+2x-3}-\frac1{x^2-1})*\frac{x^2+4x+3}{x+2}=...\\ \\ x^2+2x-3=0\\ D=2^2-4*(-3)=4+12=16\\ x_1=\frac{-2+4}{2}=1\\ x_2=\frac{-2-4}{2}=-3\\ x^2+2x-3=(x-1)(x+3)\\ \\ x^2+4x+3=x^2+2x+1+2x+2=(x+1)^2+2(x+1)=\\=(x+1)(x+1+2)=(x+1)(x+3)\\ \\ ...=(\frac{x+1}{(x-1)(x+3)}-\frac1{(x-1)(x+1)})*\frac{(x+1)(x+3)}{x+2}=\\=\frac{(x+1)^2-(x+3)}{(x-1)(x+1)(x+3)}*\frac{(x+1)(x+3)}{x+2}=\frac{x^2+2x+1-x-3}{(x-1)(x+2)}=\frac{x^2+x-2}{(x-1)(x+2)}=...$$x^2+x-2=0\\ D=1-4*(-2)=1+8=9\\ x_1=\frac{-1+3}{2}=1\\ x_2=\frac{-1-3}{2}=-2\\ x^2+x-2=(x-1)(x+2)\\ ...=\frac{(x-1)(x+2)}{(x-1)(x+2)}=1$

Answer 2

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