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bezimeni [28]
3 years ago
14

Add. 2/3 + 1/4 ...............................

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

11/12

Step-by-step explanation:

2/3+1/4

8/12+3/12= 11/12

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3 years ago
|x-3|&lt;2<br> |4x+1|&gt;0<br> |x-1|&lt;5<br> Ayudaaaaaaa pleaseeeee
Dafna11 [192]

Recuerda que

• |<em>x</em>| = <em>x</em> si <em>x</em> ≥ 0

• |<em>x</em>| = -<em>x</em> si <em>x</em> < 0

Necesitas considerar dos casos:

• si <em>x</em> - 3 ≥ 0,

|<em>x</em> - 3| < 1   ⇒  <em>x</em> - 3 < 1   ⇒   <em>x</em> < 4

• si <em>x</em> - 3 < 0,

|<em>x</em> - 3| < 1   ⇒   -(<em>x</em> - 3) = 3 - <em>x</em> < 1   ⇒   -<em>x</em> < -2   ⇒   <em>x</em> > 2

Entonces la solución consta de todos los números reales <em>x</em> tales que <em>x</em> > 2 y <em>x</em> < 4, o simplemente 2 < <em>x</em> < 4.

El método para resolver las otras desigualdades es el mismo.

|4<em>x</em> + 1| > 0   ⇒   4<em>x</em> + 1 > 0   o   -(4<em>x</em> + 1) > 0

…   ⇒   4<em>x</em> + 1 > 0   o   -4<em>x</em> - 1 > 0

…   ⇒   4<em>x</em> > -1   o   -4<em>x</em> > 1

…   ⇒   <em>x</em> > -1/4   o   <em>x</em> < -1/4

⇒   <em>x</em> ≠ -1/4

|<em>x</em> - 1| < 5   ⇒   <em>x</em> - 1 < 5   o   -(<em>x</em> - 1) < 5

…   ⇒   <em>x</em> - 1 < 5   o   -<em>x</em> + 1 < 5

…   ⇒   <em>x</em> < 6   o   -<em>x</em> < 4

…   ⇒   <em>x</em> < 6   o   <em>x</em> > -4

⇒   -4 < <em>x</em> < 6

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