Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có:

$\frac{12}{13} \times (\frac{1}{2} - \frac{1}{3}) = \frac{12}{13} \times (\frac{1 \times 3}{2 \times 3} - \frac{1 \times 2}{3 \times 2}) = \frac{12}{13} \times (\frac{3}{6} - \frac{2}{6}) = \frac{12}{13} \times \frac{1}{6} = \frac{2 \times 6}{13 \times 6} = \frac{2}{13}$

Answer 2

Ta có: x3−12=13x $x^{3} - 12 = 13 x$

⇔x3−13x−12=0 $\Leftrightarrow x^{3} - 13 x - 12 = 0$

⇔x3−x−12x−12=0 $\Leftrightarrow x^{3} - x - 12 x - 12 = 0$

⇔x(x2−1)−12(x+1)=0 $\Leftrightarrow x (x^{2} - 1) - 12 (x + 1) = 0$

⇔x(x+1)(x−1)−12(x+1)=0 $\Leftrightarrow x (x + 1) (x - 1) - 12 (x + 1) = 0$

⇔(x+1)(x2−x−12)=0 $\Leftrightarrow (x + 1) (x^{2} - x - 12) = 0$

⇔(x+1)(x2−4x+3x−12)=0 $\Leftrightarrow (x + 1) (x^{2} - 4 x + 3 x - 12) = 0$

⇔(x+1)[x(x−4)+3(x−4)]=0 $\Leftrightarrow (x + 1) [x (x - 4) + 3 (x - 4)] = 0$

⇔(x+1)(x−4)(x+3)=0 $\Leftrightarrow (x + 1) (x - 4) (x + 3) = 0$

⇔⎡⎢⎣x+1=0x−4=0x+3=0⇔⎡⎢⎣x=−1x=4x=−3 $\Leftrightarrow [\begin{matrix} x + 1 = 0 \\ x - 4 = 0 \\ x + 3 = 0 \end{matrix} \Leftrightarrow [\begin{matrix} x = - 1 \\ x = 4 \\ x = - 3 \end{matrix}$

Vậy: S={-1;4;-3}

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Answer 3