Solve, please. Quick if you can1) − x 2 + 2 x &gt; 12) | 1 − 3 x | = 53) | 4 x − 3 | ≤ 2

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Solve, please. Quick if you can1)   −      x    2    +  2  x  &amp;gt;  12)       |    1  −  3  x      |    =  53)       |    4  x  −  3      |    ≤  2

behk0 · Accepted Answer

1)   −      x    2    +  2  x  &amp;gt;  1  −      x    2    +  2  x  &amp;gt;  1  −      x    2    +  2  x  −  1  &amp;gt;  0  −  (      x    2    −  2  x  +  1  )  &amp;gt;  0  (      x    2    −  2  x  +  1  )  &amp;lt;  0  (  x  −  1      )    2    &amp;lt;  0 Square of a term is always positive, so there is no   x  ∈  R, such that (x−1)2&amp;lt;0.Hence, there is no solution2)       |    1  −  3  x      |    =  5If       |    x      |    =  a  and   a  &amp;gt;  0 , then   x  =  a or   x  =  −  a.So,       |    1  −  3  x      |    =  5 implies   1  −  3  x  =  5  or   1  −  3  x  =  −  5Then, we have:  1  −  3  x  =  5  3  x  =  1  −  5  3  x  =  −  4  x  =            −      4        3  or   1  −  3  x  =  −  5  3  x  =  1  +  5  3  x  =  6  x  =  2Therefore, the solution is   x  =  −      4    3   or   x  =  2

ttyme411gl · Accepted Answer

(3) Given:       |    4  x  −  3      |    ≤  2If       |    x      |    ≤  a and   a  &amp;gt;  0, then   −  a  ≤  x  ≤  a      |    4  x  −  3      |    ≤  2 implies   −  2  ≤  4  x  −  3  ≤  2And you get  −  2  ≤  4  x  −  3  ≤  2  −  2  ≤  4  x  −  3 and   4  x  −  3  ≤  2  4  x  ≥  −  2  +  3 and   4  x  ≤  2  +  3  4  x  ≥  1 and   4  x  ≤  5   x  ≥      1    4   and   x  ≤      5    4

Solve, please. Quick if you can 1) − x 2 </msup> + 2

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