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: 18.07.2013 19:43:26

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:
$$\frac{4 x^2-7 x+3}{3 x-1} \geq x-1$$
:
$$\frac{4 x^2-7 x+3}{3 x-1} \geq x-1\Rightarrow $$
$$\frac{4 x^2-7 x+3- \left( x-1 \right) \left( 3 x-1 \right) }{3 x-1} \geq 0$$
\(4 x^2-7 x+3- \left( x-1 \right) \left( 3 x-1 \right) \)
$$4 x^2-7 x+3- \left( x-1 \right) \left( 3 x-1 \right) = $$
:
$$4 x^2-7 x+3+ \left( -x+1 \right) \left( 3 x-1 \right) = $$
:
$$4 x^2-7 x+3-3 x^2+x+3 x-1= $$
$$x^2-3 x+2$$
: \( x^2-3 x+2 \)
\( x^2-3 x+2= 0 \)
\( x^2-3 x+2= 0 \)

.
$$D = b^2-4ac = 1$$
$$x_{1,2}= \frac{-b\pm\sqrt{D}}{2a} = \frac{3\pm\sqrt{1}}{2} = \frac{3\pm1}{2} $$
: \( x_1 = 2,\; x_2 = 1 \)
.. \( \left| a \right|=1 \), :
$$x^2+px+q=0 \Rightarrow \left\{\begin{array}{l} x_1+x_2=-p \\ x_1 \cdot x_2=q \end{array}\right.$$
$$\left\{\begin{array}{l} x_1+x_2=3 \\ x_1 \cdot x_2=2 \end{array}\right. \Rightarrow \left\{\begin{array}{l} x_1=2 \\ x_2=1 \end{array}\right.$$
: \( x_1= 2,\; x_2= 1 \)
:
$$ x_1 = 1 ;\; x_2 = 2 $$
\( 3 x-1= 0 \)
: \( x = \frac{1}{3}\)
:
  $$ \frac{1}{3} $$ $$ 1 $$ $$ 2 $$  
:
$$ x \in \left( \frac{1}{3} ;\; 1 \right] \cup \left[ 2 ;\; +\infty \right) $$
$$ \frac{1}{3} < x \leq 1 ;\;\; x \geq 2 $$


 
 
       
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