Cho A =$\frac{x^2+1}{3x}:\frac{x^2+1}{x-1}:\frac{x^3-1}{x^2+x}:\frac{x^2+2x+1}{x^2+x+1}$  và B =$\frac{x+3}{x^2-1}:\frac{x+4}{x^2+6x}-\frac{x+3}{x^2-1}:\frac{x+4}{x-4}$ . Khi x = 101, hãy so sánh A và B.

Ta có x² + (a – b)x – ab = x² + ax – bx – ab

          = x(x + a) – b(x + a) = (x – b)(x + a)

x² – (a – b)x – ab = x² – ax + bx – ab

= x(x – a) + b(x – a) = (x – a)(x + b)

x² – (a + b)x + ab = x² – ax – bx + ab

= x(x – a) – b(x – a) = (x – b)(x – a)

x² + (a + b)x + ab = x² + ax + bx + ab

= x(x + a) + b(x + a) = (x + a)(x + b)

x² – (b – 1)x – b = x² – bx + x – b

= x(x – b) + x – b = (x – b)(x + 1)

x² + (b + 1)x + b = x² + bx + x + b

= x(x + b) + x + b = (x + b)(x + 1)

x² – (b + 1)x + b = x² – bx – x + b

= x(x – b) – (x – b) = (x – b)(x – 1)

x² – (1 – b)x – b = x² – x + bx – b

= x(x – 1) + b(x – 1) = (x + b)(x – 1)

Khi đó

T= $[\frac{x^2+(a-b)x-ab}{x^2-(a-b)x-ab}.\frac{x^2-(a+b)x+ab}{x^2+(a+b)x+ab}]:[\frac{x^2-(b-1)x-b}{x^2+(b+1)x+b}.\frac{x^2-(b+1)x+b}{x^2-(1-b)x-b}]$

= $[\frac{(x-b)(x+a)}{(x-a)(x+b)}.\frac{(x-a)(x-b)}{(x+a)(x+b)}]:[\frac{(x-b)(x+1)}{(x+b)(x+1)}.\frac{(x-1)(x-b)}{(x+b)(x-1)}]$

=$\frac{{(x-b)}^2}{{(x+b)}^2}:\frac{{(x-b)}^2}{{(x+b)}^2}=1$

Vậy T = 1

Đáp án cần chọn là: A

Đáp án cần chọn là: A