if x = ,frac{4}{3}, - ,frac{{4x}}{9}, + ,,fra | vedclass

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Question

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

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$\Rightarrow x=\frac{4}{3}\left(\frac{1}{1+\frac{x}{3}}\right)=\frac{4}{3+x}$

$\Rightarrow 3 x+x^{2}=4$

$\Rightarrow x^{2}+3 x-4 \Rightarrow(x+4)(x-1)=0$

$\Rightarrow x=1,-4$

$\Rightarrow x=1$ only as $\left|-\frac{4}{3}\right|>1$

if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to

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