AMC12 2021 B

AMC12 2021 B · Q3

AMC12 2021 B · Q3. It mainly tests Manipulating equations, Fractions.

Suppose\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]What is the value of $x?$

设\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]求 $x$ 的值。

(A) \frac34 \frac34

(B) \frac78 \frac78

(D) \frac{37}{38} \frac{37}{38}

(E) \frac{52}{53} \frac{52}{53}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Subtracting $2$ from both sides and taking reciprocals gives $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$. Subtracting $1$ from both sides and taking reciprocals again gives $2+\frac{2}{3+x}=\frac{38}{15}$. Subtracting $2$ from both sides and taking reciprocals for the final time gives $\frac{x+3}{2}=\frac{15}{8}$ or $x=\frac{3}{4} \implies \boxed{\text{(A) } 3/4}$.

两边减去 $2$ 后取倒数，得 $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$。再两边减去 $1$ 后取倒数，得 $2+\frac{2}{3+x}=\frac{38}{15}$。最后两边减去 $2$ 后取倒数，得 $\frac{x+3}{2}=\frac{15}{8}$，即 $x=\frac{3}{4} \implies \boxed{\text{(A) } 3/4}$。

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