a) (x² - 1)(x² + 4x + 3) = 192

=> (x - 1)(x + 1)(x² + x + 3x + 3) - 192 = 0

=> (x - 1)(x + 1)(x + 1)(x + 3) - 192 = 0

=> [(x - 1)(x + 3)](x + 1)² - 192 = 0

=> (x² + 2x - 3)(x² + 2x + 1) - 192 = 0

Đặt x² + 2x - 3 = k

=> k(k + 4) - 192 = 0

=> k² + 4k - 192 = 0

=> k² + 16k - 12k - 192 = 0

=> k(k + 16) - 12(k + 16) = 0

=> (k - 12)(k + 16) = 0

=> (x² + 2x - 3 - 12)(x² + 2x - 3 + 16) = 0

=> (x² + 2x - 15)(x² + 2x + 13) = 0

=> x² + 5x - 3x - 15 = 0 (do x² + 2x + 13 \(\neq\)0)

=> x(x + 5) - 3(x + 5) = 0

=> (x - 3)(x + 5) = 0

=> \(\backslash\text{orbr} \left{\right. x - 3 = 0 \\ x + 5 = 0\) => \(\backslash\text{orbr} \left{\right. x = 3 \\ x = - 5\)

\( A = \frac{x^3 - 53x + 88}{(x-1)(x-3)(x-5)(x-7)+16} = \frac{(x^3 + 8x^2) - (8x^2 + 64x) + (11x+88)}{[(x-1)(x-7)][(x-3)(x-5)]+16} = \frac{x^2(x+8) - 8x(x+8) + 11(x+8)}{[x^2-8x+7][x^2-8x+15]+16} = \frac{(x+8)(x^2 - 8x + 11)}{[x^2-8x+7][x^2-8x+15]+16} \] \[ \text{Gọi } y = x^2 - 8x + 11, \quad \text{ta có: } A = \frac{(x+8)y}{(y-4)(y+4)+16} = \frac{(x+8)y}{y^2} = \frac{x+8}{y} = \frac{x+8}{x^2 - 8x + 11}. \)

\( A = \frac{x^{3} - 53x + 88}{(x-1)(x-3)(x-5)(x-7)+16} = \frac{(x^{3} + 8x^{2}) - (8x^{2} + 64x) + (11x+88)}{[(x-1)(x-7)][(x-3)(x-5)]+16} \] \[ = \frac{x^{2}(x+8) - 8x(x+8) + 11(x+8)}{[x^{2}-8x+7][x^{2}-8x+15]+16} = \frac{(x+8)(x^{2} - 8x + 11)}{[x^{2}-8x+7][x^{2}-8x+15]+16} \] Gọi \[ y = x^{2} - 8x + 11, \] ta có: \[ A = \frac{(x+8)y}{(y-4)(y+4)+16} = \frac{(x+8)y}{y^2} = \frac{x+8}{y} = \frac{x+8}{x^{2}-8x+11}. \)

\( A = \frac{x^{3} - 53x + 88}{(x-1)(x-3)(x-5)(x-7)+16} = \frac{(x^{3} + 8x^{2}) - (8x^{2} + 64x) + (11x+88)}{[(x-1)(x-7)][(x-3)(x-5)]+16} \] \[ = \frac{x^{2}(x+8) - 8x(x+8) + 11(x+8)}{[x^{2}-8x+7][x^{2}-8x+15]+16} = \frac{(x+8)(x^{2} - 8x+11)}{[x^{2}-8x+7][x^{2}-8x+15]+16} \] Gọi \(x^{2} - 8x + 11 = y\). \[ \Rightarrow A = \frac{y(x-8)}{(y-4)(y+4)+16} = \frac{y(x-8)}{y^{2}-16+16} = \frac{y(x-8)}{y^{2}} = \frac{x-8}{x^{2}-8x+11}. \)

\[

A = \frac{x^{3} - 53x + 88}{(x-1)(x-3)(x-5)(x-7)+16}

= \frac{(x^{3} + 8x^{2}) - (8x^{2} + 64x) + (11x+88)}{[(x-1)(x-7)][(x-3)(x-5)]+16}

\]

\[

= \frac{x^{2}(x+8) - 8x(x+8) + 11(x+8)}{[x^{2}-8x+7][x^{2}-8x+15]+16}

= \frac{(x+8)(x^{2} - 8x+11)}{[x^{2}-8x+7][x^{2}-8x+15]+16}

\]

Gọi \(x^{2} - 8x + 11 = y\).

\[

\Rightarrow A = \frac{y(x-8)}{(y-4)(y+4)+16}

= \frac{y(x-8)}{y^{2}-16+16}

= \frac{y(x-8)}{y^{2}}

= \frac{x-8}{x^{2}-8x+11}.

\]