# Giải bài 53 trang 58 - Sách giáo khoa Toán 8 tập 1

##### Hướng dẫn giải

a) $$1+\dfrac{1}{x}=\dfrac{x+1}{x}$$

$$1+\dfrac{1}{1+\dfrac{1}{x}}$$ = $$1+\dfrac{1}{\dfrac{x+1}{x}}$$ = $$1 + \dfrac{x}{x+1}=\dfrac{2x+1}{x+1}$$

$$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}$$ = $$1+\dfrac{1}{\dfrac{2x+1}{x+1}}$$$$1+\dfrac{x+1}{2x+1}=\dfrac{3x+2}{2x+1}$$

b) Dự đoán :

$$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}$$ = $$\dfrac{8x+5}{5x+3}$$

Thật vậy :

$$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}$$ = $$1+\dfrac{1}{1+\dfrac{1}{\dfrac{3x+2}{2x+1}}}$$ = $$1+\dfrac{1}{1+\dfrac{2x+1}{3x+2}}$$

$$1+\dfrac{1}{\dfrac{5x+3}{3x+2}}=1+\dfrac{3x+2}{5x+3}=\dfrac{8x+5}{5x+3}$$
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