Question

(|x+1|+|x-2|)/(x+199)<1

Answer 1

Ответ:(-∞;-199)∪(-66;200)

Объяснение:

Answer 2

To solve the inequality $\frac{|x+1|+|x-2|}{x+199} < 1$, we need to consider different cases based on the absolute value expressions.

Case 1: $x \geq 2$ In this case, both $x+1$ and $x-2$ are non-negative, so the absolute value expressions can be simplified as: $\frac{(x+1) + (x-2)}{x+199} < 1$ Simplify the numerator: $2x - 1 < x + 199$ Subtract $x$ from both sides: $x < 200$ However, this condition is already met in the case $x \geq 2$, so there's no additional restriction from this case.

Case 2: $1 \leq x < 2$ In this case, $x+1$ is positive, but $x-2$ is negative, so the absolute value expressions become: $\frac{(x+1) - (x-2)}{x+199} < 1$ Simplify the numerator: $3 < x + 199$ Subtract 199 from both sides: $-196 < x$ So, for this case, $x$ must satisfy $-196 < x < 2$.

Case 3: $x < 1$ In this case, both $x+1$ and $x-2$ are negative, so the absolute value expressions become: $\frac{-(x+1) - (x-2)}{x+199} < 1$ Simplify the numerator: $-2x - 1 < x + 199$ Add $2x$ to both sides: $-1 < 3x + 199$ Subtract 199 from both sides: $-200 < 3x$ Divide by 3 (remembering to reverse the inequality since we're dividing by a negative number): $x > -\frac{200}{3}$ So, for this case, $x$ must satisfy $x > -\frac{200}{3}$.

Putting all the cases together, we have the solution for the inequality: $-\frac{200}{3} < x < 2$ This is the final solution set for the given inequality.