b) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10). 20/09/2021 Bởi Sarah b) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10).
$b) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)$ $⇔ (3x – 1)(x2 + 2) – (3x – 1)(7x – 10) = 0$ $⇔ (3x – 1)(x2 + 2 – 7x + 10) = 0$ $⇔ (3x – 1)(x2 – 7x + 12) = 0$ $⇔ (3x – 1)(x2 – 4x – 3x + 12) = 0$ $⇔ (3x – 1)[(x2 – 4x) – (3x – 12)] = 0$ $⇔ (3x – 1)[x(x – 4) – 3(x – 4)] = 0$ $⇔ (3x – 1)(x – 3)(x – 4) = 0$ $⇔ 3x – 1 = 0$ hoặc $x – 3 = 0$ $+ 3x – 1 = 0 ⇔ 3x = 1 ⇔ x = 1/3.$ $+ x – 3 = 0 ⇔ x = 3.$ $+ x – 4 = 0 ⇔ x = 4.$ Vậy phương trình có tập nghiệm là $S=[$$\frac{1}{3};3;4]$ Xin CTRLHN Bình luận
`b)` `(3x-1)(x^2+2)=(3x-1)(7x-10)` `<=>(3x-1)(x^2+2)-(3x-1)(7x-10)=0` `<=>(3x-1)(x^2+2-7x+10)=0` `<=>(3x-1)(x^2-4x-3x+12)=0` `<=>(3x-1)[x(x-4)-3(x-4)]=0` `<=>(3x-1)(x-3)(x-4)=0` `<=>` \(\left[ \begin{array}{l}3x-1=0\\x-3=0\\x-4=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=3\\x=4\end{array} \right.\) Vậy phương trình trên có tập nghiệm `S={1/3;3;4}` Bình luận
$b) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)$
$⇔ (3x – 1)(x2 + 2) – (3x – 1)(7x – 10) = 0$
$⇔ (3x – 1)(x2 + 2 – 7x + 10) = 0$
$⇔ (3x – 1)(x2 – 7x + 12) = 0$
$⇔ (3x – 1)(x2 – 4x – 3x + 12) = 0$
$⇔ (3x – 1)[(x2 – 4x) – (3x – 12)] = 0$
$⇔ (3x – 1)[x(x – 4) – 3(x – 4)] = 0$
$⇔ (3x – 1)(x – 3)(x – 4) = 0$
$⇔ 3x – 1 = 0$ hoặc $x – 3 = 0$
$+ 3x – 1 = 0 ⇔ 3x = 1 ⇔ x = 1/3.$
$+ x – 3 = 0 ⇔ x = 3.$
$+ x – 4 = 0 ⇔ x = 4.$
Vậy phương trình có tập nghiệm là $S=[$$\frac{1}{3};3;4]$
Xin CTRLHN
`b)` `(3x-1)(x^2+2)=(3x-1)(7x-10)`
`<=>(3x-1)(x^2+2)-(3x-1)(7x-10)=0`
`<=>(3x-1)(x^2+2-7x+10)=0`
`<=>(3x-1)(x^2-4x-3x+12)=0`
`<=>(3x-1)[x(x-4)-3(x-4)]=0`
`<=>(3x-1)(x-3)(x-4)=0`
`<=>` \(\left[ \begin{array}{l}3x-1=0\\x-3=0\\x-4=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=3\\x=4\end{array} \right.\)
Vậy phương trình trên có tập nghiệm `S={1/3;3;4}`