Question

Sa se efectueze :
a) 9/20 - 5/36 + 7/27
b) 16/35 * 49/24 : 7/15
c) 6 intregi si 3/8 : 4 intregi si 1/4 * 10
d) 3/4 * 7,6
e) (3/5)^5 * (3/5)^4 : (2/5)^6
f) [(3 intregi si 1/2)^48]^10 * (1 intreg si 2/7)^480 totul supra (4 intregi si 1/2)^478

Answer 1

a)
9/20 - 5/36 + 7/27= 27×9-15×5+20×7 tot ăsta supra/540 =243-75+140 tot ăsta supra /540=168+140 totul supra /540=308/540 simplificat cu 2=154/270 simplificat cu2 =77/135

aflu numitorul comun al lui 20,36 si 27
20=2^2×5
36=2^2×3^2
27=3^3
------
nc=2^2×5×3^3=4×5×27=20×27=540

b) 16/35 * 49/24 : 7/15=2×7 totul supra /5×3 : 7/15=14/15×15/7=2×1 totul supra /1×1=2/1=2

c) 6 intregi si 3/8 : 4 intregi si 1/4 * 10=6×8+3 tot ăsta supra /8 : 4×4 +1 tot asta supra /4 ×10/1=51/8 : 17/4 ×10/1=51/8×4/17 ×10/1 =51×1 totul supra /2×17 × 10/1=51/34 ×10/1 =510/34 simplificat cu 2=255/17 simplificat cu 17= 15/1 =15

d) 3/4 * 7,6=3/4 × 76/10 =3×19 totul supra /1×10= 57/10

e) (3/5)^5 * (3/5)^4 : (2/5)^6 = (3/5)^(5+4) : (2/5)^6=(3/5) ^9 : (2/5)^6
cred ca ai : (3/5)^6
asa ca
3/5)^5 * (3/5)^4 : (3/5)^6 = (3/5)^(5+4) : (3/5)^6=(3/5) ^9 : (3/5)^6= (3/5)^(9-6)=(3/5)^3= 3/5×3/5×3/5=3×3×3 totul supra /5×5×5=27/125

f) [(3 intregi si 1/2)^48]^10 * (1 intreg si 2/7)^480 totul supra (4 intregi si 1/2)^478=
=[( 3×2+1 tot ăsta supra /2 )^48]^10 ×(1×7+2 tot ăsta supra /7)^480 totul supra / (4×2+1 tot ăsta supra /2)^478 =
=[(7/2)^48]^10 × (9/7)^480 totul supra / (9/2)^478=
=(7/2)^(48×10) × (9/7)^480 totul supra / (9/2)^478=
= (7/2)^480 ×(9/7)^480 totul supra /(9/2)^478=
=(9/2)^480 / (9/2)^478=
=(9/2)^480 :(9/2)^478=
=(9/2)^(480-478)=
=(9/2)^2=
daca mai trebuie socotit =>
=9/2 ×9/2 =
=81/4