`(x-5)/2-(x-3)/5=1/2`

(2)/(3)(5x-2)-(3x-(3-x)/(2))=(1-x)/(5)

(2x-1)/((x-1)(2x+3))=1/(5(x-1))-k/(5(2x+3)) , then k =

(3x-2)/(5x-1)=(3x+1)/(5x+2)

(x-1)/(3)+(2x+5)/(6)=(3x-6)/(9)-(2x-5)/(2)

Solve: (x^2-2x+5)/(3x^2-2x-5)>1/2 .

log_e ((1+3x)/(1-2x)) is equal to a) -5x-(5x^2)/2-(35x^3)/3-... b) -5x+(5x^2)/2-(35x^3)/3-... c) 5x-(5x^2)/2+(35x^3)/3-... d) 5x+(5x^2)/2+(35x^3)/3-...

2(2x-5)+3(3x-2)=5(5x-1)

Solve : (5x+2)(5x-3)(2x-1)(2x+1)=2.

((2x-1)/((x-1)(2x+3))=(1)/(5(x-1))+(k)/(5(2x+3))rArr k=

The image of the line (x-1)/3=(y-3)/1=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line (1) (x+3)/3=(y-5)/1=(z-2)/(-5) (2) (x+3)/(-3)=(y-5)/(-1)=(z+2)/5 (3) (x-3)/3=(y+5)/1=(z-2)/(-5) (3) (x-3)/(-3)=(y+5)/(-1)=(z-2)/5