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Question Number 89461 by student work last updated on 17/Apr/20

Answered by mahdi last updated on 17/Apr/20

$$\mathrm{A}=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}+\frac{\mathrm{y}}{...}}}\Rightarrow \mathrm{A}=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{A}}\Rightarrow$$$${\mathrm{A}}^2=\mathrm{Ax}+\mathrm{y}\Rightarrow {\mathrm{A}}^2-\mathrm{xA}-\mathrm{y}=0$$$$\mathrm{A}=\frac{\mathrm{x}+\sqrt{{\mathrm{x}}^2+4\mathrm{y}}}{2}\mathrm{or}\frac{\mathrm{x}-\sqrt{{\mathrm{x}}^2+4\mathrm{y}}}{2}$$

Commented by mahdi last updated on 17/Apr/20

$$\mathrm{a})6\mathrm{b})2-\sqrt{2}\mathrm{c})4\mathrm{d})4$$

Answered by $@ty@m123 last updated on 17/Apr/20

$$\left(a\right)x=5+\frac{6}{x}..\left(1\right)$$$$x^2=5x+6$$$$x^2-5x-6=0$$$$(x-6)(x+1)=0$$$$x=6,-1$$$$butx=-1isinadmissible.$$$$\therefore x=6.$$$$\left(b\right)x=3-\frac{2}{x}$$$$....$$$$....$$$$\left(c\right)x=3+\frac{4}{x}$$$$.....$$$$.....$$$$\left(d\right)x=2-\frac{8}{x}$$$$.....$$$$.....$$

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