# giúp đi mà mn oie……………….

giúp đi mà mn oie……………….

### 0 thoughts on “giúp đi mà mn oie……………….”

1. 2) $5x(2 – x) – 2 + x = 0$

$\Leftrightarrow (2-x)(5x -1) = 0$

$\Leftrightarrow \left[\begin{array}{l}2 – x = 0\\5x -1 = 0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l} x = 2\\x = \dfrac{1}{5}\end{array}\right.$

4) $x^2(2x -3) + 12 – 8x = 0$

$\Leftrightarrow x^2(2x – 3) – 4(2x -3) = 0$

$\Leftrightarrow (2x -3)(x^2 – 4) = 0$

$\Leftrightarrow (2x – 3)(x-2)(x +2) = 0$

$\Leftrightarrow \left[\begin{array}{l}2x -3 = 0\\x -2 = 0\\x +2 = 0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x= \dfrac{3}{2}\\x =2\\x= -2\end{array}\right.$

6) $x^3 +3x^2 +3x +1 = 8x^3$

$\Leftrightarrow (x +1)^3 = (2x)^3$

$\Leftrightarrow x + 1 = 2x$

$\Leftrightarrow x = 1$

8) $x^3 + 2x^2 = -x$

$\Leftrightarrow x(x^2 + 2x +1) = 0$

$\Leftrightarrow x(x +1)^2 = 0$

$\Leftrightarrow \left[\begin{array}{l}x = 0\\x + 1 = 0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = 0\\x = -1\end{array}\right.$

2. $2) 5x(2 – x) – 2 + x = 0$

$⇔ 5x(2 – x) – (2 – x) = 0$

$⇔ (2 – x)(5x – 1) = 0$

$⇔ \left[ \begin{array}{l}2-x=0\\5x-1=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=2\\x=\frac{1}{5}\end{array} \right.$

$4) x²(2x – 3) + 12 – 8x = 0$

$⇔ x²(2x – 3) – (8x – 12) = 0$

$⇔ x²(2x – 3) – 4(2x – 3) = 0$

$⇔ (x² – 4)(2x – 3) = 0$

$⇔ (x – 2)(x + 2)(2x – 3) = 0$

$⇔\left[\begin{array}{ccc}x-2=0\\x+2=0\\2x-3=0\end{array}\right]$

$⇔\left[\begin{array}{ccc}x=2\\x=-2\\x=\frac{3}{2}\end{array}\right]$

$6) x³ + 3x² + 3x + 1 = 8x³$

$⇔ 7x² – 3x² – 3x – 1 = 0$

$⇔ (x – 1)(7x² + 4x + 1) = 0$

$⇔ x – 1 = 0$

$⇔ x = 1$

Giải thích:

Ta có: $7x² + 4x + 1 = 7x² + 4x + \frac{4}{7} + \frac{3}{7} = (x\sqrt{7} + \frac{2\sqrt{7}}{7})² + \frac{3}{7} > 0$ với mọi x => $7x² + 4x + 1 = 0$ là vô lý ( vô nghiệm )

$8) x³ + 2x² = -x$

$⇔ x³ + 2x² + x = 0$

$⇔ x(x² + 2x + 1) = 0$

$⇔ x(x + 1)² = 0$⇔\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\$