+752 \(x = {1+\frac{\sqrt{2}}{1+\frac{\sqrt{2}}{1+...}}}\). Find \(\frac{1}{(x+1)(x-2)}\). When your answer is in the form \(\frac{A+\sqrt{B}}{C}\) , where A, B, and C are integers, what is |A|+|B|+|C|?
+118608 \(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).
\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)
\(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \) .
When your answer is in the form \(\frac{A+\sqrt{B}}{C}\) ,
A=+2, B=+2, C= -2
where A, B, and C are integers, what is |A|+|B|+|C|?
|A|+|B|+|C| = 2+2+2 = 6
+118608 \(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).
\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)
\(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \) .
When your answer is in the form \(\frac{A+\sqrt{B}}{C}\) ,
A=+2, B=+2, C= -2
where A, B, and C are integers, what is |A|+|B|+|C|?
|A|+|B|+|C| = 2+2+2 = 6
