题目已知a1=x+1(x≠0,a_{1}=x+1(x≠0,a1=x+1(x=0,且x≠−1)x≠-1)x=−1)，a2=11−a1a_{2}=\dfrac{1}{1-a_{1}}a2=1−a11，a3=11−a2a_{3}=\dfrac{1}{1-a_{2}}a3=1−a21，…，an=11−an−1.a_{n}=\dfrac{1}{1-a_{n-1}}.an=1−an−11.若a2022a_{2022}a2022的值为202220222022，则xxx的值为 ______.知识点：解一元一次方程章节：方程与不等式 / 一元一次方程
