Equation and Inequation With Absolute Value [12th Grade]

shawqidu19
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Homework Statement



Resolve these equations and these inequations with the absolute values. Give the solutions in the form of interval :

|2-x|< 4

|6-2x| = 3

|x+2| > 3

|4x²-12x+9| = 4

|3x+1|+|1-x|>3

|1-x²|=2x

|x+2|<|x+3|

|x^3-1|+pi&amp;gt;\sqrt{3}&lt;br /&gt; &lt;br /&gt; 3&amp;lt;|x+2|&amp;lt;4&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;h2&gt;Homework Equations&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;
 
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Way too hard to solve without some hints.
Have you done some work to help me out with?
 
you should try do them yourselves, I'm sure that if you think about them you can solve them =).

to get a 'feel'

|y|<c if and only if

-y>-c ánd y<c

if there is one you get stuck on, I or someone else will help you as you've shown what you've tried.
 
I have always felt that the best way to handle complicated inequalities is to first solve the connected equation (|2- x|= 4, etc.) The points where the two sides are equal separate "<" and ">".
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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