Question

八年级数学 已知（2x+y﹚﹙2x+y-1）=2，求x+1/2y的值

Answer 1

解令t=2x+y
则由（2x+y﹚﹙2x+y-1）=2
得t（t-1）=2
即t²-t-2=0
即（t-2)(t+1）=0
解得t=2或t=-1
即2x+y=2或2x+y=-1
即2（x+1/2y）=2或2（x+1/2y）=-1
即x+1/2y=1或x+1/2y=-1/2

Answer 2

（2x+y﹚﹙2x+y-1）=2
（2x+y)²-(2x+y)-2=0
(2x+y-2)(2x+y+1)=0
∴2x+y-2=0或2x+y+1=0
∴x+1/2y=1或x+1/2y=-1/2

Answer 3

解，两边同时除以4，得：（x+y/2）（x+y/2-1/2）=1/2，得：（x+y/2）²-1/2（x+y/2）=1/4，设a=（x+y/2）=a，得：a²-a/2=1/4，两边同时减去1/4，得：a²-a/2+1/4=1/2，（a-1/4）²=1/2，a=√2/2+1/4，即（x+y/2）=√2/2+1/4

Answer 4

把2x+y看成一个整体作为z，z*(z-1)=2,得到2x+y为2或-1，所以x+1/2y为1或-1/2