Question

Rút gọn biểu thức :

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

Giúp tôi với

Answer 1

Ta có:

\(A=\frac{x^3-53x+88}{\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)+16}=\frac{\left(x^3+8x^2\right)-\left(8x^2+64x\right)+\left(11x+88\right)}{\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]+16}=\frac{x^2\left(x+8\right)-8x\left(x+8\right)+11\left(x+8\right)}{\left[x^2-8x+7\right]\left[x^2-8x+15\right]+16}=\frac{\left(x+8\right)\left(x^2-8x+11\right)}{\left[x^2-8x+7\right]\left[x^2-8x+15\right]+16}\)

Gọi \(x^2-8x+11=y\)

\(\Rightarrow A=\frac{y\left(x-8\right)}{\left(y-4\right)\left(y+4\right)+16}=\frac{y\left(x-8\right)}{y^2-16+16}=\frac{y\left(x-8\right)}{y^2}=\frac{x-8}{x^2-8x+11}\)

 

Answer 2

\[

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}.

\]

Answer 3

\( 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}. \)

Answer 4

\( 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}. \)

Answer 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}Gọiy=x^2-8x+11,tacó: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}.\\)

Answer 6

\( 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}. \)

Answer 7

\( 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}. \)

Answer 8

Answer 9

Answer 10