I. 6. (-1/4)⁵/(-1/4)²·(-1/4)⁻²=
(-1/4)³·(-1/4)⁻²=
-1/4
II. 1. Notam
a/3=b/4=c/5=d/6=k
a=3k
b=4k
c=5k
d=6k
a+b+c+d=360(de grade) sau
3k+4k+5k+6k=18k=360
k=20
a=60(de grade-primul unghi);
b=80(de grade-al doilea unghi);
c=100(de grade-al treilea unghi);
d=120(de grade-al patrulea unghi);
2. 2x+3 /7 -x/2=3x-6 /14
4x+6-7x /14=3x-6 /14
4x+6-7x=3x-6
-3x+6=3x-6
-3x-3x=-6-6
-6x=-12
-x=-2 ⇒x=2
3. (a+b+c)/3=17 ⇒a+b+c=51
(b+c)/2=16 ⇒b+c=32
c=3b ⇒b+3b=4b=32
b=8
c=24
a=19
4. a. A romb=d·D/2
A romb=24cm²
d·D=48cm²(produsul diagonalelor);
b. A romb= b·h /2 +b·h /2
A romb= 2·b·h/2
A romb=b·h
A romb=24cm²
b=8cm ⇒h=3cm(inaltimea rombului);
III. 1. 6/25+[28/25/(5/3·12/5-2/3]=
6/25+[28/25/(4-2/3)]=
6/25+(28/25/10/3)=
6/25+42/125=
30+42 /125=72/125
2. Avem
ABCD-trapez isoscel unde AB║CD si AB<CD
AB=5cm ,AD=BC=6cm si m(<DCB)=60 de grade;
a. Construim AE⊥DC ,E∈(DC) si BF⊥DC ,F∈(DC) de unde obtinem
AEFB-dreptunghi
m(<E)=m(<F)=90 de grade) ⇒AB=EF=5cm
De asemenea avem ΔAED≡BFC (m(<D)=m(<F)=60 de grade AD=BC) .
Obtinem ca cateta care se opune unghiului de 30 de grade este jumatate din lungimea ipotenuzei deci
DE=CF=3cm
CD=DE+EF+FC
CD=3cm+5cm+3cm=11cm(lungimea segmenului CD);
b. Folosind teorema asemanarii obtinem
ΔMAB≈ΔMDC ⇔
MA/MD=MB/MC=AB/DC=5/11
MA/MA+AD=5/11
MA/MA+6=5/11
MA=5cm
MB/MB+BC=5/11
MB/MB+6=5/11
MB=5cm
P MCD=MD+DC+MC
P MCD=11cm+11cm+11cm
P MCD=33cm(perimetrul ΔMCD)
