ĐKXĐ : x khác 2 ; -2 ; 1
Ta có : A = (x-2/x-1 - x+3/2-x - 3x^2 - 5x + 3/x^2 - 3x + 2 ) : x + 2 / x - 1
= [x-2/x-1 + x + 3 / x - 2 - 3x^2 - 5x + 3 / (x-1)(x-2)] . x-1/x+2
= [(x-2)^2 / (x-2)(x-1) + (x+3)(x-1) / (x-1)(x-2) - 3x^2 - 5x + 3 / (x-1)(x-2) ] . x-1/x+2
= (x-2)^2 + (x+3)(x-1) - 3x^2 + 5x - 3 / (x-1)(x-2) . x-1/x+2
= x^2 - 4x + 4 + x^2 + 2x - 3 - 3x^2 + 5x - 3 / (x-1)(x-2) . x-1/x+2
= -x^2 + 3x - 2 / (x-1)(x-2) . x-1/x+2
= -(x^2 - 3x + 2) / (x-1)(x-2) . x-1/x+2
= -(x-1)(x-2)/(x-1)(x-2) . x-1/x+2
= -1 . x - 1/x+2
= 1-x/x+2
P = A . (x^2 + 5/1-x)
=1-x/x+2 . x^2 + 5 / 1 - x
= x^2 + 5 / x + 2
= (x^2 - 4 + 9 )/ x + 2
= (x-2) + 9/x+2
= x+2 + 9/x+2 - 4
Do x >= 0 => x + 2 >= 2 > 0
Áp dụng BĐT Cô - si , ta có :
[tex]x + 2 + \frac{9}{x+2} \geq 2\sqrt{(x+2).\frac{9}{x+2}} = 6[/tex]
=> x + 2 + 9/x+2 - 4 >= 2
Dấu " = " xảy ra <=> x + 2 = 9/x+2
<=> (x+2)^2 = 9
<=> [tex]\begin{bmatrix} x+2=3 & \\ x+2=-3 & \end{bmatrix}[/tex]
<=> [tex]\begin{bmatrix} x=1(L) & \\ x=-5(l) & \end{bmatrix}[/tex]
=> Sai đề