Question

pun in joc ultimele puncte 22.
a)Fie E(x)=1/10×(1/2x-5-1/2x+5):x²+3x+2/4x²-25 unde x e R /{-5/2; -2; -1; 5/2} aratati ca E(x)=1/(x+1)(x+2)
b)Calculati E(0)+E(1)+E(2)+..........+E(2007).
Si ultima:
ABCDA'B'C'D' un cub cu muchia de 4 cm. Punctele M si N sunt mijloacele laturilor A'D' si respectiv D'C'.
a) calculati lungimea segmentului MN;
b) calculati aria trapezului MNCA;
c) calculati sinusul unghiului dintre dreptele AM si DN.
Imi pare rau ca nu mai am puncte

Answer 1

a) E(x) = 1/10 x [(2x+5-2x+5)/(2x+5)(2x-5)] : (x+2)(x+1)/(2x-5)(2x+5)
E(x) = 1/10 x 10/(2x+5)(2x-5) x (2x+5)(2x-5)/(x+2)(x+1)
E(x)= 1/1 x 1/1 x 1/(x+2)(x+1)
E(x)= 1/(x+1)(x+2)
b) 1/1x2  + 1/2x3 + 1/3x4+...+1/2008 x 2009
=2-1/1x2  + 3-2/2x3  +4-3/3x4+....+2009-2008/2009x2008
=2/1x2 -1/1x2 +3/2x3-2/2x3+....+2009/2008x2009 - 2008/2008x2009
=1/1-1/2+1/2-1/3+...+1/2008-1/2009
=1/1-1/2009
=2009/2009 - 1/2009
=2008/2009

a) in ΔA'D'C' - MN linie mijlocie⇒ MN=A'C' :2
A'C'-diagonala in patrat⇒ A'C'=l√2=4√2 ⇒ MN=4√2:2=2√2 cm

b) trapezul MNCA - isoscel si AM=NC
in Δ MA'A dr⇒ TP⇒ AM²= A'M² + A'A²= 4²+2²=16+4=20
AM=√20= 2√5 cm=NC si AC=4√2cm
H trapezului AMNC : H²=(2√5)²-√2²=20-2=18⇒ H=3√2
A AMNC= (MN+AC) * H :2= (2√2+4√2) * 3√2 :2= 6√2*3√2:2=18*2:2=18cm²

c) AM=DN=2√5
ducem AE || DN ( in patratul AA'B'B), ME=MN=2√2 si AE=DN
ΔAME isoscel ,∡(AM,DN)= ∡ (AM,AE)=∡MAE
hΔMAE = H MNCA= 3√2
AΔAME= 3√2 * 2√2:2=6*2:2=6cm²
h2 Δ MAE= AΔAME *2 :AM=6*2:2√5=6/√5=6√5/5
sin∡MAE= 6√5/5 :2√5=3/5
sin∡(AM;DN)= 3/5