Đáp án:
Giải thích các bước giải:
$\eqalign{ & {x^3} - 8{x^2} = \frac{x}{4} - 2 \cr & \Leftrightarrow {x^2}(x - 8) = \frac{{x - 8}}{4} \cr & \Leftrightarrow {x^2}(x - 8) - \frac{{x - 8}}{4} = 0 \cr & \Leftrightarrow (x - 8)({x^2} - \frac{1}{4}) = 0 \cr & \Leftrightarrow (x - 8)(x - \frac{1}{2})(x + \frac{1}{2}) = 0 \cr & \Leftrightarrow x - 8 = 0\,hoặc\,x - \frac{1}{2} = 0\,hoặc\,x + \frac{1}{2} = 0 \cr & \Leftrightarrow x = 8\,hoặc\,x = \frac{1}{2}\,hoặc\,x = - \frac{1}{2} \cr} $
$\eqalign{ & \frac{{2x}}{{x - 5}} - 3{x^2} + \frac{{9x}}{{{x^2} - 25}} + \frac{{x - 9}}{{x + 7}} \cr & = \frac{{2x(x + 5)}}{{(x - 5)(x + 5)}} - \frac{{3{x^2}(x - 5)(x + 5)}}{{(x - 5)(x + 5)}} + \frac{{9x}}{{(x - 5)(x + 5)}} + \frac{{x - 9}}{{x + 7}} \cr & = \frac{{2{x^2} + 10x - 3{x^2}({x^2} - 25) + 9x}}{{(x - 5)(x + 5)}} + \frac{{x - 9}}{{x + 7}} \cr & = \frac{{ - 3{x^4} + 77{x^2} + 19x}}{{(x - 5)(x + 5)}} + \frac{{x - 9}}{{x + 7}} \cr & = \frac{{( - 3{x^4} + 77{x^2} + 19x)(x + 7) + (x - 9)({x^2} - 25)}}{{(x - 5)(x + 5)(x + 7)}} \cr & = \frac{{ - 3{x^5} - 21{x^4} + 78{x^3} + 549{x^2} + 108x + 225}}{{(x - 5)(x + 5)(x + 7)}} \cr} $
$\eqalign{ & A = {a^3} + {b^3} \cr & = {(a + b)^3} - 3ab(a + b) \cr & = {3^3} - 3.( - 10).3 \cr & = 117 \cr} $
