Middle school math problems, which require a detailed process to be written. 5

There's a lot of detail on this (and it's already perfect).

1) Solution: ab=cd, the reason is: pass o as oe ab in e, of cd in f, connect ob, od, apm= cpm, apm= bpn, cpm= dpn, bpn= dpn, oe ab, of cd, oe=of, in rt beo and rt dof, of=oe, od=ob, obtained by the pythagorean theorem:

be=df,of cd,oe ab,of,oe over o, obtained by the perpendicular diameter theorem: cd=2df,ab=2be,ab=cd

2) ab=cd is true, proving: o is oe ab in e, of cd is in f, ob, od, apm= cpm, oe=of, in rt beo and rt dof, of=oe, od=ob, from the pythagorean theorem: be=df, of cd, oe ab, of, oe over o, from the perpendicular diameter theorem:

cd=2df，ab=2be，ab=cd．

Over O, as oe ab and e point, of cd and f point, (auxiliary line) pn bisector bpd, oe=of (the distance from the point on the angle bisector to both sides of the angle is equal) ab=cd (in the circle, the chord centric distance is equal, then the chord is equal).

Both questions can be solved in this way.

Solution: Set the time of 1 episode of TV series to A, the time of 1 sketch to B, and the time of a song to C(B>C).

There is 4a+2b=4b+6c=the capacity of a disc.

2a=b+3c

14a+22b+40c=7(b+3c)+22b+40c=29b+61c=7(4b+6c)+b+19c

2(4b+6c)-(b+19c)=7b-7cb>c 2(4b+6c)-(b+19c)>0, so 7+2=9 discs are needed.

As can be seen from the title, 4 TV dramas = 2 sketches + 6 songs, 2 TV dramas = 1 sketch + 3 songs.

Extrapolated from 14 TV series, 12 TV series + 8 sketches = 3 discs.

The remaining 2 TV series can be converted to get 2 TV series = 1 sketch + 3 songs, plus 3 sketches, 3 songs get 1 CD.

Remaining sketches = 22-8-3 = 11, remaining songs = 40-3 = 37

Because of the "Sketch Time" song time, 11 sketches + 1 song + 18 songs = 3 discs.

The remaining songs are 37-19=18

Based on 1 disc with 10 songs, 2 more are needed.

A minimum of 3 + 1 + 3 + 2 = 9 discs is required.

Let the time of 1 episode of TV series be A, the time of 1 sketch is B, and the time of a song is C4A+2B=4B+6C

2a=b+3c

14a+22b+40c=7(b+3c)+22b+40c=29b+61c=7(4b+6c)+b+19c

2(4b+6c)-(b+19c)=7b-7cb>c

2(4b+6c)-(b+19c)>0

So 7+2=9 discs are needed.

A is a positive number B is a negative number C is a negative number Can be pushed out |b+c|=-b-c，|c-a|=a-c, which can be reduced to: |a+b|-|c-a|-|b+c|=|a+b|-（a-c）-（b-c）=|a+b|-a+c+b+c=|a+b|-a+b+2c

If |a|Greater than |b|, then |a+b|=a+b, then |a+b|-|c-a|-|b+c|=2b+2c

If |a|Less than |b|, then |a+b|=-b-a, then |a+b|-|c-a|-|b+c|=-2a+2c

A is a positive number, B is a negative number, and c is a negative number.

c-a<0 b+c<0

When a>-b, the original formula = a+b+c-a+b+c=2(b+c) When a<=-b, the original formula =-a-b+c-a+b+c=2c-2a

Let PBQ be similar to ABC.

There are two scenarios.

ba=bq:bc

8-2t):8=4t:16

t=:bc=bq:ba

8-2t):16=4t:8

t = 4 3 A or 4 3 seconds later.

Hope it helps.

Let's say: After X seconds, BP is 8-2X, BQ is 4XpBQ is similar to ABC.

So bp ab=bq bc

When AB = 8 cm, BC = 16 cm, BP is 8-2x, and BQ is 4X.

x=2A2 after two seconds.

But... If QBP is similar to ABC, this is a different answer.

bp/ac=bq/bc

When AB = 8 cm, BC = 16 cm, BP is 8-2x, and BQ is 4X.

x=4 3A: 4 after 3s.

When PBQ is similar to ABC, :bp:ab=bo:bc,(8-2t):8=4t:16,t=2

bp:bc=bo:ab，(8-2t):16=4t:8，t=4/5

So t=2 or 4 5

It is calculated using the factorization method.

x-2m）*（x-1/2）=0

x1=1/2

x2 = 2m, then 2m = root number 3 2

m = root number 3 4