解:据题意得:1<a<2
∴ 原式\(\displaystyle =a-1+\left| {a-2} \right|\)
\(\displaystyle =a-1-a+2\)
\(\displaystyle =1\)
例题2:已知实数a,b在数轴上表示的点如图所示,试化简下列各式
(1)、\(\displaystyle \sqrt{{{{a}^{2}}}}+\left| {ab} \right|\)
(2)、\(\displaystyle \left| {a+b} \right|+\left| {a-b} \right|\)
解析:通过图示可以看出a,b互为相反数
解:
(1)、原式\(\displaystyle =-a-ab\)
\(\displaystyle =-a+{{a}^{2}}\)
(2)、原式\(\displaystyle =0+b-a\)
\(\displaystyle =-2a\)
原式\(\displaystyle =\left| a \right|-\left| {a+c} \right|+\left| {c-b} \right|-\left| {-b} \right|\)
\(\displaystyle =-a+a+c-c+b-b\)
\(\displaystyle =-a+a+c-c+b-b\)
