The math problems in the third year of junior high school require a solution process

Connecting BM Because AB is the diameter, the angle AMB is a right angle.

It is derived from the intersecting string theorem.

ap*pm=bp*pn

ap*am+bp*bn

ap*(ap+pm)+bp*(bp+pn)ap^2+ap*pm+bp*pn+bp^2ap^2+2ap*pm+pm^2+(bp^2-pm^2)am^2+bm^2ab^2

ap*am+bp*bn

ap*(ap+pm)+bp*(bp+pn)ap 2+ap*pm+bp 2+bp*pn, according to the intersecting chord theorem.

ap*pm=bp*pn

So, the original formula = ap 2 + bp 2 + 2 ap * pm connects to bm, obviously the angle amb is the rt angle, bp 2 = bm 2 + pm 2 + 2 ap * pm + pm 2 + bm 2 = (ap + pm) 2 + bm 2 = am 2 + bm 2 = 36

Connect BM, AN

Because ab is the diameter, amb= anb=90°

It is derived from the intersecting string theorem.

ap*pm=bp*pn, so ap*am+bp*bn

ap*(ap+pm)+bp*(bp+pn)ap^2+ap*pm+bp*pn+bp^2ap^2+2ap*pm+pm^2+(bp^2-pm^2)am^2+bm^2 ab^2

It's like a concrete lazy thought made with the intersecting string theorem.

A question comes:

1.If the equation for x x 2+2(a+1)x+(a 2+4a-5)=0 has a real root, try to find it.

The value of a positive integer.

So =0

That is, [2(a+1)] 2-4(a 2+4a-5) = 0 solution: a = 3

It is advisable to do the math again.

1.If the equation for x x 2+2(a+1)x+(a 2+4a-5)=0 has a real root, try to find the value of the positive integer a.

Solution: The equation has a real root, so: (2(a+1)) 2-4(a 2+4a-5)>=0

Just solve it.

Green is followed by blue-purple, yellow is preceded by red, and the middle is a continuous 5, so a total of 5n+3 (n is a positive integer), and the end is either 3 or 8.

Because green is followed by blue, with blue as the first position of the cycle, according to the red, yellow-green, blue-purple 5-color cycle, the middle interception is a multiple of 5 minus 2 (yellow-green), so only 2008 meets the conditions in the given answer

Solution: Set a price reduction of X yuan.

The equation is (20+5x)(44-x)=1600 because x>0 so x=4

Answer: The price will be reduced by 4 yuan.

Set a price reduction of x yuan.

1600=44(5x+20)

Then solve the equation.

Solution: (1) Set the motion time to t

When POA=90°, 90 12 180=2 t or (360-90) 12 180=2 t

Solution: t=3 or 9

2) Link OP, PA and PB

poa×12×π÷180=2π×2

poa=60°

po=ao poa is an equilateral triangle.

Pa=OA and AB=OA

pa=oa=ab

b=∠apb

b=∠apb=30

opb=90°

op is the radius.

The straight line bp is tangent to o.

Kneel and beg to adopt ......Thank you).

When POA=90°, the arc PA=quarter-circumference=6, so time=62=3s.

When the motion time is 2s, the arc PA=4 , and the corresponding central angle is 60°, and OA=OP, OAP is an equilateral triangle, AB=OA=AP, and OPB is a right triangle, that is, OPB=90°

pb is the tangent of o.