Question

假设 $2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+x}}}=\dfrac{144}{53}\text{。}$

$x$ 的值是多少？

Suppose $2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+x}}}=\dfrac{144}{53}.$

What is the value of $x?$

$\dfrac{3}{4}$

$\dfrac{7}{8}$

$\dfrac{14}{15}$

$\dfrac{37}{38}$

$\dfrac{52}{53}$

从外向内一层一层剥开，先从 $\dfrac{144}{53}$ 中减去 $2$ 。

Peel off one layer at a time, starting by subtracting $2$ from $\dfrac{144}{53}$

在 $\dfrac{144}{53}-2=\dfrac{38}{53}$ 后，取倒数并继续向内化简

After $\dfrac{144}{53}-2=\dfrac{38}{53},$ take reciprocals and continue inward

从外向内计算， $\dfrac{144}{53}-2=\dfrac{38}{53}$ ，所以内层分式等于 $\dfrac{38}{53}$ 。

取倒数得 $1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38}$ ，所以 $\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}$ 。

于是 $2+\dfrac{2}{3+x}=\dfrac{38}{15}$ ，所以 $\dfrac{2}{3+x}=\dfrac{8}{15}$ ，得到 $3+x=\dfrac{15}{4}$ 。

因此 $x=\dfrac{15}{4}-3=\dfrac{3}{4}$ 。

所以正确答案是 A。

Working from the outside in, $\dfrac{144}{53}-2=\dfrac{38}{53},$ so the inner fraction equals $\dfrac{38}{53}.$

Its reciprocal gives $1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38},$ so $\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}.$

Then $2+\dfrac{2}{3+x}=\dfrac{38}{15},$ so $\dfrac{2}{3+x}=\dfrac{8}{15},$ giving $3+x=\dfrac{15}{4}.$

Therefore $x=\dfrac{15}{4}-3=\dfrac{3}{4}.$

Thus, the correct answer is A.