解析:观察算法框图发现:

当x>1时,有y=x-2;

当x≤1时,有y=2x,

所以y={■(2^x "," x≤1"," @x"-" 2"," x>1"." )┤

答案:y={■(2^x "," x≤1"," @x"-" 2"," x>1)┤

7.如图所示的算法框图表示求函数y=|x-3|的值.请将算法框图补充完整.其中①处应填　　　　　　　　　,②处应填　　　　　.

答案:x<3(或x≤3)　x-3

8.已知算法框图如图所示.

若输出的是 1/4,则输入的是__________________.

解析:由算法框图知y={■(2^("-" x) "," x≤1"," @log_81 x"," x>1"." )┤

由y=1/4,知当2-x=1/4 时,x=2,与x≤1不符,舍去;

当log81x=1/4 时,x=3,满足x>1.故输入的是3.

答案:3

9.如果学生的成绩大于或等于60分,那么输出"及格",否则输出"不及格".画出解决该问题的算法框图.