$\frac{1}{x}$ + $\frac{1}{}$ $\sqrt[2]{10-x^2}$ =4/3

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1. $\dfrac{1}{x}+\dfrac{1}{\sqrt{10-x²}}=\dfrac{4}{3}$ $(x\neq0;x²\neq10)$

   $⇔\dfrac{1}{\sqrt{10-x²}}=\dfrac{4x-3}{3x}$ $(-\sqrt{10}<x<\sqrt{10})$

   $⇔\dfrac{1}{10-x²}=\dfrac{16x²-24x+9}{9x²}$

   $⇔9x²=(10-x²)(16x²-24x+9)$

   $⇔9x²=160x²-240x+90-16x^{4}+24x³-9x²$

   $⇔16x^{4}-24x³-142x²+240x-90=0$

   $⇔8x^{4}-12x³-71x²+120x-45=0$

   …

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