Question

Ex 1,2,3 si 4 va rog mult​

Answer 1

1. a) A = {-5; -4; -3; -2; -1; 0; 1}

b) B= {-3}

c) C= {-3; -2; -1; 0; 1; 2; 3}

d) D= {-2; -1; 0; 1; 2; 3; 4}

2. {2,3,4,5,6} d.p.{a,b,c,d,e}

a/2=b/3=c/4=d/5=e/6=k

a=2k

b=3k

c=4k

d=5k

e=6k

2k+3k+4k+5k+6k=1200

20k=1200

k=1200÷20

k=60

a=60×2=120

b=60×3=180

c=60×4=240

d=60×5=300

e=60×6=360

3. {a,b,c} i.p. {2,3,4,}

a²+b²+c²=976

2a=3b=4c

2a=k (coeficient de proporționalitate)⇒a=$\frac{k}{2}$

3b=k ⇒b=$\frac{k}{3}$

4c=k ⇒c=$\frac{k}{4}$

()²+()²+()²=976

$\frac{k^{2} }{4}$+$\frac{k^{2} }{9}$+$\frac{k^{2} }{16}$=976

++=976

=976

61k²=144*976

61k²=140544

k²=2304

k=48

a==24

b==16

c=$\frac{48}{4}$=12

4. a) 3+(-2)-5= 1-5= -4

b) 6-(-2)*(-3)= 8*(-3)= -2

c) 5:(-5)+(-5)= -1+(-5)=-6

d) $(-2)^{3} -(-2)^{2} =$-8-(-4)= -12

e) I-3I+I2I =

i. -3+2= -1

ii. -3-2 = -5

iii. 3 + 2 = 5

iv. 3-2 = 1

f) I-5I-I-7I=

i. -5-(-7)= 2

ii. 5-(-7)= 12

iii. -5-7= -12

iv. 5-7= -2

g) I$(-2)^{2}$I - I$(-2)^{3}$I =

I4I-I-8I=

i. 4-(-8)=12

ii. 4-8= -4

iii. -4 -(-8)= 4

iv. -4-8 = -12

h) II-4I-I6II=

i. I-4-6I = I-10I => a. -10 ; b. 10

ii. I4-6I = I-2I => a. -2 ; b. 2

iii. I-4-(-6)I = I2I => a. 2 ; b. -2

iv. I4-(-6)I = I10I => a. 10 ; b. -10

i) -1+3*{-3-2*[-1-(5-3)]^2

-1+3*[-3-2*(-1-2)]^2

-1+3*(9+36-18) -- am folosit formula de calcul prescurtat (a-b)^2

=1+81= 82

j) {[(-3)^2+1^2-2*(-3)]+(-12)+(-8)+1}^3

[(9+1-6)+(-12)+(-8)+1]^3

(4-12-8+1)^3

4^3+12^3+8^3+1^3+3*(4+(-12))*(4+(-8))*4*((-12)*(-8))*(-12)*(-8)

sau

-15^3 = -3375