Question

1 Daca nr. x si y sunt invers proportionale cu 6 si 5,sa se arate ca $\frac{4x-y}{2x+3y} = \frac{1}{2}$
2 Daca numerele x si y sunt invers proportionale cu 3 si 4 aflati valoarea rapoartelor: $\frac{x+2y}{6x-5y} ; \frac{6x-5y}{2x+y} ; \frac{ x^{2}+ y^{2} }{ x^{2} +xy} ; 2x^{2} +3xy+ y^{2} \frac{2x^{2} +3xy+ y^{2}}{ x^{2} +xy +2 y^{2} }$
Va rog,ajuta-ti-ma...Multumesc!

Answer 1

6x=5y
x=5y/6
(4x-y)/(2x+3y)=(4×5y/6-y)/[(2×5x/(6+3y)=(10y-3y)/3 ×3/(5y+9y)=7y//14y=1/2

3x=4y
x=4y/3
(x+2y)/6x-5y)=(4y/3+2y)/(6×4y/3-5y)=(4y+6y)/3 × 3/(24y-15y)=10y/9y=10/9

(6x-5y)/(2x+y) =6×4y/3-5y)/(2×4y/3+y)=(24y-5y)/3 × 3/(8y+3y)= 9y/11y=9/11

(x²+y²)/(x²+xy) =[(4y/3)²+y²]/[(4y/3)²+4y/3×y]=(16y²+9y²)/9 × 9/(16y²+12y²)=25y²/28y²=25/28

(2x²+3xy+y²)/(x²+xy+2y²)=[2×(4y/3)²+3×4y/3×y+y²]/[(4y/3)²+4y/3×y+2y²)=

=(32y²+36y²+9y²)/9 × 9/(16y²+12y²+18y²)=77y²/46y²=77/46