Question

Problema 1 dar daca stiti si primul exercitiu va rog ajutatima dar imi trebuie problema mai mult dau coroana

Answer 1

(x²-81)/(x³+18x²+81x)=
=(x-9)(x+9)/x(x²+18x+81)=
=(x-9)(x+9)/x(x+9)²=
=(x-9)/x(x+9)


1.
CM=MB=40:2=20
a)DM=√(40²+20²)=√(1600+400)=√2000=20√5

b)
triunghiul DMN dreptunghic in M
ducem MP perpendicular pe AD
ipotenuza =DA+AN
DP=20
PN=20+x
teorema inaltimii
MP²=20×(20+x)
40²=400+20x
1600=400+20x
1200=20x
x=60

c)
caz 1. cand DN=AD+AN=AD+x=40+x
daca x=30 m
triunghiul DMN are baza DN=40+30=70    si inaltimea MP=40
A=70×40/2=1400 m²
Apatrat=40×40=1600 m²
1400/1600=14/16=7/8=0,875=87,5%

caz 2. cand DN=DN+x=40     DN=40-30=10
A=10×40/2=200 m²
Apatrat=40×40=1600 m²
200/1600=1/8=0,125=12,50%

Answer 2

(x-9)(x+9)/x(x²+18+81)=
(x-9)(x+9)/x(x+9)²=
(x-9)/x(x+9)

a)DM,il,aflam,direct,din,ΔDCM,dr.
Stim,ca,CM=20(M,este,mijl.lui,BC),CD=40
=>DM²=CD²+CM²
DM²=400+1600=>DM²=2000=>DM=20√5

b)In,tr.dr.DAN,AD²+AN²=DN²;1600+X²=DN²=2000+MN²(ΔDMN,dr-ipoteza)
In,tr,dr,NBM,NB²+BM²=MN²;(40-X)²+400=MN²
1600+X²=2000+(40-X)²+400
1600+X²=2000+1600-80X+X²+400
2400=80X=>X=30

c)A=1600m²
A(ΔDMN)=10√5x20√5/2=500m²

1600__________100%
500___________a
a=50000/1600=>a=31,25%