Question

Rezozolvati inecuatiile
1) lg(3x−17)=lg(x+1)
2) log2 (x - 1) - log2 (2x - 4) = 0
3)lg(x ² - 2x) = lg(2x + 12)
4)log2 (2x ² + 7) = log2 (x ² + 8)
5)log2 (x² + 3) = log2 (2x ² + x + 1)
6)lg(2x ^ 2 + 3x) = lg(6x + 2)​

Answer 1

Explicație pas cu pas:

pentru fiecare ecuație, stabilim domeniul de definiție

1.

$3x - 17 > 0;x + 1 > 0 = > x > \frac{17}{3} \\ 3x - 17 = x + 1 \\ 3x - x = 1 + 17\\ 2x = 18 = > x = 9$

2.

$x - 1 > 0;2x - 4 > 0 = > x > 2 \\ x - 1 = 2x - 4 \\ x - 2x = - 4 + 1 \\ - x = - 3 = > x = 3$

3.

$x ² - 2x > 0;2x + 12 > 0 \\= > - 6 < x < 0; x > 2\\ x ² - 2x = 2x + 12 \\ x ² - 4x - 12 = 0 \\ (x + 2)(x - 6) = 0 \\ x = - 2;x = 6$

4.

$2x ² + 7 > 0;x ² + 8 > 0 = > x \: real\\ 2x ² + 7 = x ² + 8 \\ 2 {x}^{2} - {x}^{2} = 8 - 7 \\ {x}^{2} = 1 \\ x = - 1;x = 1$

5.

$x² + 3 > 0; 2x ² + x + 1 > 0 = > x \: real \\ x² + 3 = 2x ² + x + 1 \\ x² + 3- 2x ² - x - 1 = 0 \\ - {x}^{2} - x + 2 = 0 \\ {x}^{2} + x - 2 = 0 \\ (x + 2)(x - 1) = 0 \\ x = - 2;x = 1$

6.

$2 {x}^{2} + 3x > 0;6x + 2 > 0 = > x > 0 \\ 2 {x}^{2} + 3x = 6x + 2 \\ 2 {x}^{2} - 3x - 2 = 0 \\ (2x + 1)(x - 2) = 0 = > x = - \frac{1}{2};x = 2 \\ = > x = 2$