Question

Решить неравенство : х+1/3-х>1

Answer 1

№1 Решите неравенство
x-1/3+x+2/4>5x-38/2.

№2 Решите систему неравенств:

система 3x меньше или равно-5 система -3x<4
4x>-7; -2xбольше или равно 3;


система 2-3x<11+2x
1-2x<6+x

Answer 2

$x + \dfrac{1}{3-x}> 1 \\ \\ ODZ: 3-x\ne 0 \ ; \ x \ne 3 \\ \\ \dfrac{x(3-x)+1-3+x}{3-x} > 0 \\ \\ \dfrac{3x-x^{2} - 2 + x}{3-x} > 0 \ \ / *(-1) \\ \\ \dfrac{x^{2} - 4x + 2}{3-x}< 0 \\ \\ \dfrac{(x-2+\sqrt{2})(x-2-\sqrt{2})}{3-x}< 0 \ (1) \\ \\ x \in (2-\sqrt{2};3)\cup (2+\sqrt{2} ; +\infty)$

Ответ: x ∈ (2-√2; 3)∪(2+√2; + ∞)

Answer 3

To solve the inequality $x + \frac{1}{3} - x > 1$, we need to simplify the expression and find the range of values for $x$ that satisfy the inequality.

Step 1: Combine like terms on the left-hand side of the inequality:

$x + \frac{1}{3} - x$ simplifies to $\frac{1}{3}$.

Step 2: The inequality now becomes:

$\frac{1}{3} > 1$.

However, there seems to be a mistake in the inequality. The inequality $\frac{1}{3} > 1$ is not true since $\frac{1}{3}$ is less than 1. Therefore, the inequality in the original question is not correct.

If you meant a different inequality or if there was a typo, please provide the correct inequality, and I'll be happy to help you solve it.