Question

1. (q-2,6b) tengsizliklar sistemasini yeching: \\(\

$$\begin{cases}5x - 18 < 27 \\\\ x + 5 > 15\\end{cases}$$

\\)
a) \\(x \in (10; \infty)\\)
b) \\(x \in (-\infty; 9)\\)
c) \\(x \in \emptyset\\)
d) \\(x \in (-\infty; -10)\\)

1. (q-2,6b) tengsizliklar sistemasini yeching: \\(\

$$\begin{cases}3x - 8 > 5 \\\\ x^2 - 3x + 2 < 0\\end{cases}$$

\\)
a) \\(x \in (-\infty; -1)\\)
b) \\(x \in \emptyset\\)
c) \\(x \in (-1; 3)\\)
d) \\(x \in (1; 2)\\)

1. (q-2,6b) tenglamani yeching: \\(\sqrt{x^2 + x - 2} = \sqrt{1 + 3x}\\)

a) \\(-1\\)
b) \\(3\\) va \\(-1\\)
c) \\(-2; 1\\) va \\(-\frac{1}{3}\\)
d) \\(3\\)

1. (q-2,6b) tenglamani yeching: \\((x^2 - 25) \cdot \sqrt{x - 3} = 0\\)

a) \\(-5\\) va \\(3\\)
b) \\(-5; 3\\) va \\(5\\)
c) \\(3\\) va \\(5\\)
d) \\(-5\\) va \\(5\\)

Explanation:

Question 11

Step1: Solve first inequality

Solve \(5x - 18 < 27\). Add 18 to both sides: \(5x < 27 + 18 = 45\). Divide by 5: \(x < 9\).

Step2: Solve second inequality

Solve \(x + 5 > 15\). Subtract 5: \(x > 10\).

Step3: Find intersection

The first inequality gives \(x < 9\), the second \(x > 10\). No overlap, so solution set is empty.

Step1: Solve first inequality

Solve \(3x - 8 > 5\). Add 8: \(3x > 13\), so \(x > \frac{13}{3} \approx 4.33\).

Step2: Solve second inequality

Solve \(x^2 - 3x + 2 < 0\). Factor: \((x - 1)(x - 2) < 0\). The solution is \(1 < x < 2\).

Step3: Find intersection

First inequality: \(x > \frac{13}{3}\), second: \(1 < x < 2\). No overlap, so solution set is empty.

Step1: Square both sides

Square both sides of \(\sqrt{x^2 + x - 2} = \sqrt{1 + 3x}\) to get \(x^2 + x - 2 = 1 + 3x\).

Step2: Simplify equation

Rearrange: \(x^2 - 2x - 3 = 0\). Factor: \((x - 3)(x + 1) = 0\). Solutions: \(x = 3\) or \(x = -1\).

Step3: Check domain

For \(\sqrt{x^2 + x - 2}\), \(x^2 + x - 2 \geq 0\) → \((x + 2)(x - 1) \geq 0\) → \(x \leq -2\) or \(x \geq 1\). For \(\sqrt{1 + 3x}\), \(1 + 3x \geq 0\) → \(x \geq -\frac{1}{3}\).

• Check \(x = 3\): \(3 \geq 1\) and \(3 \geq -\frac{1}{3}\) → valid.
• Check \(x = -1\): \(-1 \geq -\frac{1}{3}\) is false → invalid.

Answer:

C) \(x \in \emptyset\)

Question 12