IV $$ 1^{\circ} $$ Expand : $$ (x-1)(2 x+5) $$ and $$ (x+1)(x-2) $$ $$ 2^{\circ} $$ Solve $$ : \frac{5 x-1}{x^{2}-x-2} \geqslant-2 $$ $$ 3^{\circ} $$ Solve : $$ |1-2 x| \leqslant|x-2| $$ $$ 4^{\circ} $$ Deduce the solution of the system : $$ \left\{\begin{array}{l}\frac{2 x^{2}+3 x-5}{x^{2}-x-2} \geqslant 0 \ |1-2 x| \leqslant|x-2|\end{array} ight. $$

Question

IV $$ 1^{\circ} $$ Expand : $$ (x-1)(2 x+5) $$ and $$ (x+1)(x-2) $$ $$ 2^{\circ} $$ Solve $$ : \frac{5 x-1}{x^{2}-x-2} \geqslant-2 $$ $$ 3^{\circ} $$ Solve : $$ |1-2 x| \leqslant|x-2| $$ $$ 4^{\circ} $$ Deduce the solution of the system : $$ \left\{\begin{array}{l}\frac{2 x^{2}+3 x-5}{x^{2}-x-2} \geqslant 0 \ |1-2 x| \leqslant|x-2|\end{array}ight. $$

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Let's tackle each part of the problem step by step. ### Part 1: Expand $$ (x-1)(2x+5) $$ and $$ (x+1)(x-2) $$ 1. **Expand $$ (x-1)(2x+5) $$**: $$ (x-1)(2x+5) = x \cdot 2x + x \cdot 5 - 1 \cdot 2x - 1 \cdot 5 $$ $$ = 2x^2 + 5x - 2x - 5 = 2x^2 + 3x - 5 $$ 2. **Expand $$ (x+1)(x-2) $$**: $$ (x+1)(x-2) = x \cdot x + x \cdot (-2) + 1 \cdot x + 1 \cdot (-2) $$ $$ = x^2 - 2x + x - 2 = x^2 - x - 2 $$ ### Part 2: Solve $$ \frac{5x-1}{x^{2}-x-2} \geqslant -2 $$ First, we need to rewrite the inequality: $$ \frac{5x-1}{x^{2}-x-2} + 2 \geqslant 0 $$ This can be combined into a single fraction: $$ \frac{5x-1 + 2(x^2-x-2)}{x^2-x-2} \geqslant 0 $$ Now, let's simplify the numerator: $$ 5x - 1 + 2x^2 - 2x - 4 = 2x^2 + 3x - 5 $$ So the inequality becomes: $$ \frac{2x^2 + 3x - 5}{x^2 - x - 2} \geqslant 0 $$ Next, we will find the roots of the numerator and denominator to analyze the sign of the expression. 1. **Roots of the numerator $$ 2x^2 + 3x - 5 = 0 $$**: Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} $$ $$ = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4} $$ This gives us: $$ x_1 = 1 \quad \text{and} \quad x_2 = -\frac{5}{2} $$ 2. **Roots of the denominator $$ x^2 - x - 2 = 0 $$**: $$ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} $$ This gives us: $$ x_3 = 2 \quad \text{and} \quad x_4 = -1 $$ Now we have the critical points: $$ -\frac{5}{2}, -1, 1, 2 $$. We will analyze the sign of the expression in the intervals defined by these points. ### Part 3: Solve $$ |1-2x| \leqslant |x-2| $$ To solve this absolute value inequality, we will consider the cases based on the critical points where the expressions inside the absolute values change sign. 1. **Case 1**: $$ 1 - 2x \geq 0 $$ and $$ x - 2 \geq 0 $$ (i.e., $$ x \leq \frac{1}{2} $$ and $$ x \geq 2 $$ - no solutions). 2. **Case 2**: $$ 1 - 2x \geq 0 $$ and $$ x - 2 < 0 $$ (i.e., $$ x \leq \frac{1}{2} $$ and $$ x < 2 $$): $$ 1 - 2x \leq 2 - x \implies 1 \leq 2 - x + 2x \implies 1 \leq 2 + x \implies x \geq -1 $$ Thus, $$ -1 \leq x \leq \frac{1}{2} $$. 3. **Case 3**: $$ 1 - 2x < 0 $$ and $$ x - 2 \geq 0 $$ (i.e., $$ x > \frac{1}{2} $$ and $$ x \geq 2 $$): $$ -1 + 2x \leq x - 2 \implies 2x - x \leq -2 + 1 \implies x \leq -1 \quad \text{(no solutions)} $$ 4. **Case 4**: $$ 1 - 2x < 0 $$ and $$ x - 2 < 0 $$ (i.e., $$ x > \frac{1}{2} $$ and $$ x < 2 $$): $$ -1 + 2x \leq 2 - x \implies 3x \leq 3 \implies x \leq 1 $$ Thus, $$ \frac{1}{2} < x < 1 $$. Combining the results from all cases, we have: $$ [-1, \frac{1}{2}] \cup (\frac{1}{2}, 1] $$ ### Part 4: Deduce the solution of the system Now we need to combine the results from Part 2 and Part 3. 1. From Part 2, we have the intervals from the inequality $$ \frac{2x^2 + 3x - 5}{x^2 - x - 2} \geqslant 0 $$. 2. From Part 3, we have the intervals $$ [-1, \frac{1}{2}] \cup (\frac{1}{2}, 1] $$. We will find the intersection of these intervals to deduce the solution of the system. Now, let's compute the intervals for Part 2 and find the intersection with the intervals from Part 3 To solve the system: 1. **Expand the expressions**: - $$ (x-1)(2x+5) = 2x^2 + 3x - 5 $$ - $$ (x+1)(x-2) = x^2 - x - 2 $$ 2. **Solve the inequality** $$ \frac{5x-1}{x^{2}-x-2} \geqslant -2 $$: - Rewrite as $$ \frac{5x-1 + 2(x^2 - x - 2)}{x^2 - x - 2} \geqslant 0 $$ - Simplify to $$ \frac{2x^2 + 3x - 5}{x^2 - x - 2} \geqslant 0 $$ - Find critical points: $$ x = 1, -\frac{5}{2}, 2, -1 $$ - Analyze intervals and determine where the expression is non-negative. 3. **Solve the absolute value inequality** $$ |1-2x| \leqslant |x-2| $$: - Break into cases based on the critical points $$ x = \frac{1}{2} $$ and $$ x = 2 $$ - Find the solution intervals: $$ [-1, \frac{1}{2}] \cup (\frac{1}{2}, 1] $$ 4. **Combine the solutions**: - Find the intersection of the intervals from Part 2 and Part 3 to get the final solution for the system. **Final Solution**: The solution to the system is the intersection of the intervals from the inequality and the absolute value inequality, resulting in $$ x $$ values that satisfy both conditions.

IV Expand : and
Solve
Solve :
Deduce the solution of the system :

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Part 1: Expand and

Part 2: Solve

Part 3: Solve

Part 4: Deduce the solution of the system

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