a=b=c
Then
4a^2+36 a^2> or =K*a^3/(3a)
Then k< or =120
Then
Maximum value forK is 120
...
\[\begin{gathered}
Day:\,\,196.\,\,\,\,\,\,\,\,\,\,\,Solve\,\,in\,\,\ mathbb{R}\,\,\,\,\,the\,\,eqaution\,\,\frac{{2x}}
{{3{x^2} - x + 2}} - \frac{{7x}}
{{3{x^2} + 5x + 2}} = 1 \hfill \\
Sol\,\,\,\,\frac{{2x}}
{{3{x^2} - x + 2}} - \frac{{7x}}
{{3{x^2} + 5x + 2}} = 1 & & \div \frac{x}
{x}\,\,\,,x \ne 0 \hfill \\
\,\,\,\, \Rightarrow \frac{2}
{{3x - 1 + \frac{2}
{x}}} - \frac{7}
{{3x + 5 + \frac{2}
{x}}} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,\,3x + \frac{2}
{x} = u \hfill \\
\,\,\,\, \Rightarrow \frac{2}
{{u - 1}} - \frac{7}
{{u + 5}} = 1\,\, \hfill \\
\,\,\,\, \Rightarrow 2u + 10 - 7u + 7 = {u^2} + 4u - 5 \hfill \\
\,\,\,\, \Rightarrow {u^2} + 9u - 22 = 0\,\,\,\,\,\, \Rightarrow \left( {u + 11} \right)\left( {u - 2} \right) = 0\,\,\,\, \Rightarrow Either\,\,u = - 11\,\,Or\,\,u = 2\, \hfill \\
1.\,\,\,\,\,If\,\,\,u = - 11\,\,\,\,\, \Rightarrow \,3x + \frac{2}
{x} = - 11\,\,\,\, \Rightarrow 3{x^2} + 11x + 2 = 0\,\,;\,\,{b^2} - 4ac = 121 - 24 = 97 \hfill \\
\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}
{{2a}} = \, = \frac{{ - 11 \pm \sqrt {97} }}
{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ ,\,\, \hfill \\
\,2.\,\,If\,\,\,\,u = 2 \Rightarrow 3x + \frac{2}
{x} = 2\,\,\,\, \Rightarrow 3{x^2} - 2x + 2 = 0\,\,\,\,\,\,\,;\,\,{b^2} - 4ac = 4 - 24 = - 24 < 0 \hfill \\
No\,\,solution\,\,in\,\,\mathbb{R}\, \hfill \\
\therefore \,\,\,\,\,\,\,\,\,S.S. = \left\{ {\frac{{ - 11 + \sqrt {97} }}
{6},\frac{{ - 11 - \sqrt {97} }}
{6}} \right\} \hfill \\
\end{gathered} \]
; 02-17-2018 12:54 PM
: 1 (0 1 )
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