Question

1. (1/x-1) + 2/x > (1/x+1) 2. (x+1)(x+2)(x+3)/(x-1)(x-2)(x-3)>1

Answer 1

$1)\; \; \frac{1}{x-1}+\frac{2}{x}>\frac{1}{x+1}\\\\\frac{x(x+1)+2(x-1)(x+1)-x(x-1)}{x(x-1)(x+1)}>0\\\\\frac{x^2+x+2x^2-2-x^2+x}{x(x-1)(x+1)}>0\\\\\frac{2(x^2+x-1)}{x(x-1)(x+1)} >0\\\\x^2+x-1=0\; ,\; \; D=1+4=5\; ,\; x_{1,2}=\frac{-1\pm \sqrt5}{2}\\\\x_1\approx -1,62\; ,\; \; x_2\approx 0,62\\\\---( -1,62)+++(-1)---(0)+++(0,62)---(1)+++\\\\x\in (\frac{-1-\sqrt5}{2},-1)\cup (0,\frac{-1+\sqrt5}{2})\cup (1,+\infty )$

$2)\; \; \frac{(x+1)(x+2)(x+3)}{(x-1)(x-2)(x-3)}>1\\\\\frac{(x+1)(x+2)(x+3)-(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}>0\\\\\frac{(x^2+3x+2)(x+3)-(x^2-3x+2)(x-3)}{(x-1)(x-2)(x-3)}>0\\\\\frac{x^3+3x^2+3x^2+9x+2x+6-(x^3-3x^2-3x^2+9x+2x-6)}{(x-1)(x-2)(x-3)}>0\\\\\frac{12x^2+12}{(x-1)(x-2)(x-3)}>0\; ,\; \; \frac{12(x^2+1)}{(x-1)(x-2)(x-3)}>0\\\\12(x^2+1)>0\; \; pri\; \; x\in (-\infty ,+\infty )\; \; \Rightarrow (x-1)(x-2)(x-3)>0\\\\---(1)+++(2)---(3)+++\\\\x\in (1,2)\cup (3,+\infty )$

Answer 2

To solve these inequalities, we need to find the values of 'x' that satisfy them. Let's go through each one step by step:

1. (1/x - 1) + 2/x > (1/x + 1)

To simplify the inequality, we'll first find a common denominator for all the fractions. The common denominator is 'x'.

[(1 - x)/x + (2/x)] > [(1 + x)/x]

Now, let's clear the fractions by multiplying both sides by 'x':

(x)(1 - x) + 2 > (x)(1 + x)

Expand the expressions:

x - x^2 + 2 > x + x^2

Now, bring all terms to one side of the inequality:

x - x^2 + 2 - x - x^2 > 0

Combine like terms:

-2x^2 + x + 2 > 0

Now, let's find the critical points by setting the expression equal to zero:

-2x^2 + x + 2 = 0

To solve this quadratic equation, you can use the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

In this case, a = -2, b = 1, and c = 2:

x = [-(1) ± √(1^2 - 4(-2)(2))] / 2(-2) x = [-1 ± √(1 + 16)] / -4 x = [-1 ± √17] / -4

So, the critical points are x = (-1 + √17)/-4 and x = (-1 - √17)/-4.

Now, let's analyze the intervals formed by these critical points and test them in the original inequality:

Interval 1: x < (-1 - √17)/-4 Interval 2: (-1 - √17)/-4 < x < (-1 + √17)/-4 Interval 3: x > (-1 + √17)/-4

Pick a test point from each interval and evaluate the inequality to determine whether it holds true.

For Interval 1, let's take x = -2: -2 - (-2)^2 + 2 > -2 + (-2)^2 -2 - 4 + 2 > -2 + 4 -4 > 2

This is false, so Interval 1 is not part of the solution.

For Interval 2, let's take x = 0: 0 - (0)^2 + 2 > 0 + (0)^2 2 > 0

This is true, so Interval 2 is part of the solution.

For Interval 3, let's take x = 2: 2 - (2)^2 + 2 > 2 + (2)^2 2 - 4 + 2 > 2 + 4 0 > 6

This is false, so Interval 3 is not part of the solution.

Therefore, the solution to the inequality is: (-1 - √17)/-4 < x < (-1 + √17)/-4

However, please note that dividing both sides of the inequality by 'x' could introduce extraneous solutions, so it's essential to check the solutions we obtained against the original inequality to confirm their validity. In this case, we found that only Interval 2 is valid.

Now, let's move on to the second inequality:

2. (x + 1)(x + 2)(x + 3)/(x - 1)(x - 2)(x - 3) > 1

First, let's simplify the expression:

[(x + 1)(x + 2)(x + 3)] / [(x - 1)(x - 2)(x - 3)] > 1

Now, let's find the critical points by setting the expression equal to 1:

(x + 1)(x + 2)(x + 3) = (x - 1)(x - 2)(x - 3)

Expand both sides:

x^3 + 6x^2 + 11x + 6 = x^3 - 6x^2 + 11x - 6

The x^3 terms cancel out, and we are left with:

6x^2 + 6 = -6x^2 - 6

Combine like terms:

12x^2 + 6 = 0

Now, divide both sides by 6:

2x^2 + 1 = 0

Subtract 1 from both sides:

2x^2 = -1

Finally, divide by 2:

x^2 = -1/2

This equation has no real solutions since you cannot take the square root of a negative number in the real number system. Thus, there are no critical points for this inequality.

Now, let's analyze the expression's behavior for large positive and negative values of 'x':

As 'x' becomes very large (positive or negative), the numerator and denominator both grow as the cube of 'x', which means their ratio approaches 1.

Thus, the inequality holds true for all real values of 'x' except at x = ±√(1/2), where it is undefined.

The solution to the inequality is: x ∈ (-∞, -√(1/2)) U (-√(1/2), +∞)

Again, keep in mind the importance of checking the validity of solutions, especially when working with fractions and expressions involving 'x'. Always verify that the solutions satisfy the original inequality to ensure their correctness.