: A = (15Vx - 11) /( x + 2Vx - 3) + (3Vx - 2) / (1 -Vx) - ( 2Vx + 3 )/ ( Vx + 3)
Nếu vậy thì: A = ( 15Vx - 11) / (Vx - 1)( Vx + 3) - ( 3 Vx -2)( Vx+3) /( Vx-1)(Vx+3) -(2Vx+3)(Vx-1) / (Vx-1)(Vx+3) ĐK: x # 1 , x>=0
= [ 15Vx - 11 - (3Vx-2)(Vx+3) - (2Vx+3)( Vx - 1) ] / ( Vx-1)(Vx+3)
= ( 15Vx - 11 - 3x -9Vx + 2Vx +6 - 2x + 2Vx - 3Vx + 3 ) / ( Vx-1)(Vx+3)
= ( 7Vx -5x -2 ) / (Vx-1(Vx+3) = ( -5x + 5Vx +2Vx - 2) /( (Vx-1)(Vx + 3)
= (Vx-1)( 2 -5Vx) / (Vx-1)(Vx+3) = (2- 5Vx) / (Vx+3)
b) A= (2 - 5Vx)/ ( Vx+3) = (- 5 Vx - 15 + 17) /( Vx+3) = [ -5( Vx + 3) + 17 ] / (Vx +3)
= -5 + 17 /( Vx +3)
Vì Vx >= 0 với mọi x
=> Vx + 3 >= 3 (Dấu ''='' xảy ra <=> x=0 )
=> 17 /( Vx+3) <= 17/3 => -5 + 17/(Vx+3) <= 2/3
=> A=2/3 <=> x=0
Vậy Max= 2/3 <=> x=0
c) Để A = 1/2 thì ( 2 - 5Vx)(Vx+3) = 1/2 <=> 2(2 - 5Vx) = Vx+3
<=> 4 - 10Vx - Vx - 3 = 0 <=> -11Vx = -1 <=> Vx =1 => x =1 (KTMĐK *)
Vậy không có giá tị nào của x để A=1/2
d)A= (2 - 5Vx)/ ( Vx+3) = (- 5 Vx - 15 + 17) /( Vx+3) = [ -5( Vx + 3) + 17 ] / (Vx +3)
= -5 + 17 /( Vx +3)
Vì Vx >= 0 với mọi x
=> Vx + 3 >= 3
=> 17 /( Vx+3) <= 17/3 => -5 + 17/(Vx+3) <= 2/3
=> A <= 2/3 ( Dấu ''='' xảy ra <=> x=0)