Задача 66905 ...

62aaac79225ec1501a9067b6

16.11.2022 12:41:45

([m]\frac{x+sqrt(x²-4)+2}{x-sqrt(x²-4)+2}[/m]+[m]\frac{2-sqrt(x²-4)+x}{2+sqrt(x²-4)+x}[/m])x^-1.

626fab9a354ab65fb2f43abb

16.11.2022 15:44:34

★

[m](\frac{x+\sqrt{x^2-4}+2}{x-\sqrt{x^2-4}+2}+\frac{2-\sqrt{x^2-4}+x}{2+\sqrt{x^2-4}+x})x^{-1}=(\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}} + \frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}})*\frac{1}{x}= [/m]
[m]= \frac{(x+2+\sqrt{x^2-4})^2 + (x+2-\sqrt{x^2-4})^2}{(x+2-\sqrt{x^2-4})(x+2+\sqrt{x^2-4})}*\frac{1}{x}= [/m]
[m] =\frac{(x+2)^2 + 2(x+2)\sqrt{x^2-4} + x^2-4 + (x+2)^2-2(x+2)\sqrt{x^2-4}+x^2-4}{(x+2)^2-(x^2-4)}*\frac{1}{x}=[/m]
[m]= \frac{2(x^2+4x+4) + 2x^2-8}{x^2+4x+4-x^2+4}*\frac{1}{x} =\frac{2x^2+8x+8 + 2x^2-8}{4x+8}*\frac{1}{x}=\frac{4x^2+8x}{4x+8}*\frac{1}{x}=1[/m]