Question

Ex. 5 (si 6 daca se poate)

Answer 1

Acesta este exercitiu 5 a. Daca nu intelegi spune.mi!

Answer 2

b)  =[7x/2(x+2)]-[(3x+5)/3(x+1)]+[(2-3x)/((x+1)]-[(x+6)/6(x+1)]
=[7x·3-2(3x+5)+6(2-3x)-(x+6)]/6(x+1)
=(21x-6x-10+12-18x-x-6)/6(x+1)
=(-4-4x)/6(x+1)
=-4(x+1)/6(x+1)
=-4/6

6)a) =[4/(x+2)]-[(x+10)/(x-2)(x+2)]+[3/(x-2)]
=[4(x-2)-x-10+3(x+2)]/(x+2)(x-2)
=(4x-8-x-10+3x+6)/(x+2)(x-2)
=(6x-12)/(x+2)(x-2)
=6(x-2)/(x+2)(x-2)
=6/(x+2)

b) =[(7x-1)/2x(x+3)]-[(3x-5)/(x²-9)]+[3(3x-1)/2x(x²-9)]
=[(x-3)(7x-1)-2x(3x-5)+3(3x-1)]/[2x(x²-9)]
=(7x²-x-21x+3-6x²+10x+9x-3)/2x(x+3)(x-3)
=(x²-3x)/2x(x+3)(x-3)
=x(x-3)/2x(x+3)(x-3)
=1/2(x+3)

c) =[(x+1)/x(x-1)]-[(x+2)/2(x²-1)]+[1/-2x(x²-1)]
=[(x+1)/x(x-1)]-[(x+2)/2(x-1)(x+1)]+[-1/2x(x-1)(x+1)]
=[2(x+1)(x+1)-x(x+2)-1]/2x(x+1)(x-1)
=(2x²+4x+2-x²-2x-1)/2x(x+1)(x-1)
=(x²+2x+1)/2x(x+1)(x-1)
=(x+1)²/2x(x+1)(x-1)
=(x+1)/2x(x-1)

d) =[3/2(x-4)]+[(3-2x)/2x(x+4)]-[(15x-12)/x(x²-16)]
=[3x(x+4)+(3-2x)(x-4)]/[2x(x+4)(x-4)]-[2(15x-12)/x(x-4)(x+4)]
=(3x²+12x+3x-12-2x²+8x-30x+24)/2x(x+4)(x-4)
=(x²-7x+12)/2x(x+4)(x-4)
=(x-3)(x-4)/2x(x+4)(x-4)
=(x-3)/2x(x+4)