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 ...