Question

Даны 2 неравенства: 1) x-1 <или равно √5 2) 5-2x>5-2√17 Изобразите решение каждого неравенства на числовой прямой и найдите множество значений x, не удовлетворяющих ни одному из неравенств

Answer 1

1) x-1≤√5
x≤1+√5, √5≈2.24,   x≤3.24
2) x<√17,  √17≈4.12
   x>√17

Answer 2

1.x-1≤√5
  5-2x>5-2√17,
 x≤1+√5,         x≤√5+1
-2x>-2√17,      x<√17.
Т.к. √17>√5+1,то все числа х≥√17 не удовлетворяют ни одному из неравенств

Answer 3

Solution to the First Inequality: $x - 1 \leq \sqrt{5}$

To solve the first inequality, we need to isolate the variable $x$ on one side of the inequality sign. Let's go through the steps:

1. Add 1 to both sides of the inequality: $x - 1 + 1 \leq \sqrt{5} + 1$ Simplifying, we get: $x \leq \sqrt{5} + 1$

Now, let's plot the solution on a number line. Since $x$ is less than or equal to $\sqrt{5} + 1$, we will shade the number line to the left of $\sqrt{5} + 1$ and include the endpoint:

<---|-------------------|--->
   0   √5 + 1

Solution to the Second Inequality: $5 - 2x > 5 - 2\sqrt{17}$

To solve the second inequality, we'll follow a similar process:

1. Subtract 5 from both sides of the inequality: $5 - 2x - 5 > 5 - 5 - 2\sqrt{17}$ Simplifying, we get: $-2x > -2\sqrt{17}$

2. Divide both sides of the inequality by -2. Since we are dividing by a negative number, the inequality sign will flip: $\frac{-2x}{-2} < \frac{-2\sqrt{17}}{-2}$ Simplifying, we get: $x < \sqrt{17}$

Now, let's plot the solution on a number line. Since $x$ is less than $\sqrt{17}$, we will shade the number line to the left of $\sqrt{17}$:

<---|-------------------|--->
   0   √17

Intersection of the Solutions

To find the values of $x$ that satisfy both inequalities, we need to find the intersection of the shaded regions on the number line. From the previous plots, we can see that the shaded region for the first inequality is to the left of $\sqrt{5} + 1$, and the shaded region for the second inequality is to the left of $\sqrt{17}$. Therefore, the values of $x$ that satisfy both inequalities are the values to the left of both $\sqrt{5} + 1$ and $\sqrt{17}$.

<---|-------------------|--->
   0   √5 + 1    √17

In interval notation, the solution set is $(-\infty, \sqrt{5} + 1) \cap (-\infty, \sqrt{17})$, which means all values of $x$ less than $\sqrt{5} + 1$ and less than $\sqrt{17}$ satisfy both inequalities.