题目
若f(x)=1+x (x)=dfrac (2-x)(x-1)，则f(x)=1+x (x)=dfrac (2-x)(x-1)f(x)=1+x (x)=dfrac (2-x)(x-1) f(x)=1+x (x)=dfrac (2-x)(x-1) f(x)=1+x (x)=dfrac (2-x)(x-1)f(x)=1+x (x)=dfrac (2-x)(x-1)
若
，则
题目解答
答案
由题得
因为
所以
故本题答案选
解析
步骤 1：计算$f(\dfrac {1}{2})$
根据函数$f(x)=1+x$，将$x=\dfrac {1}{2}$代入，得到$f(\dfrac {1}{2})=1+\dfrac {1}{2}=\dfrac {3}{2}$。
步骤 2：计算$g(f(\dfrac {1}{2}))$
根据函数$g(x)=\dfrac {2-x}{x-1}$，将$x=f(\dfrac {1}{2})=\dfrac {3}{2}$代入，得到$g(f(\dfrac {1}{2}))=g(\dfrac {3}{2})=\dfrac {2-\dfrac {3}{2}}{\dfrac {3}{2}-1}=\dfrac {\dfrac {1}{2}}{\dfrac {1}{2}}=1$。
根据函数$f(x)=1+x$，将$x=\dfrac {1}{2}$代入，得到$f(\dfrac {1}{2})=1+\dfrac {1}{2}=\dfrac {3}{2}$。
步骤 2：计算$g(f(\dfrac {1}{2}))$
根据函数$g(x)=\dfrac {2-x}{x-1}$，将$x=f(\dfrac {1}{2})=\dfrac {3}{2}$代入，得到$g(f(\dfrac {1}{2}))=g(\dfrac {3}{2})=\dfrac {2-\dfrac {3}{2}}{\dfrac {3}{2}-1}=\dfrac {\dfrac {1}{2}}{\dfrac {1}{2}}=1$。