1.
Connect AQ,
AQ=BD=3
Because angle ADQ= angle BDC
AQ is parallel to BC.
So triangle ADQ is similar to triangle CDB.
So ad /DC = AQ/ BC
So AD/(6-AD)=3/5.
Calculate the answer yourself. ....
2。
DE=AC-AD-CE
According to the algorithm of the first question, we can get
AD=6X/(5+X)
Similarly, according to similar triangles (self-certification)
AC/BC=CE/PC
6/5=CE/(5-X)
CE=6-6/5X
So Y=6-6X/(5+X)-(6-6/5X)
Y=6X/(5+X)+6X/5
If it can be simplified ... Do it yourself = =
The third question is considered according to the situation. The method is not much more than the previous two questions. It's similar ... I won't do it, brother. Come on!
Good luck in the senior high school entrance examination ...