Toán $\frac{1}{x}$ + $\frac{1}{}$ $\sqrt[2]{10-x^2}$ =4/3 20/09/2021 By Daisy $\frac{1}{x}$ + $\frac{1}{}$ $\sqrt[2]{10-x^2}$ =4/3
$\dfrac{1}{x}+\dfrac{1}{\sqrt{10-x²}}=\dfrac{4}{3}$ $(x\neq0;x²\neq10)$ $⇔\dfrac{1}{\sqrt{10-x²}}=\dfrac{4x-3}{3x}$ $(-\sqrt{10}<x<\sqrt{10})$ $⇔\dfrac{1}{10-x²}=\dfrac{16x²-24x+9}{9x²}$ $⇔9x²=(10-x²)(16x²-24x+9)$ $⇔9x²=160x²-240x+90-16x^{4}+24x³-9x²$ $⇔16x^{4}-24x³-142x²+240x-90=0$ $⇔8x^{4}-12x³-71x²+120x-45=0$ … Trả lời
$\dfrac{1}{x}+\dfrac{1}{\sqrt{10-x²}}=\dfrac{4}{3}$ $(x\neq0;x²\neq10)$
$⇔\dfrac{1}{\sqrt{10-x²}}=\dfrac{4x-3}{3x}$ $(-\sqrt{10}<x<\sqrt{10})$
$⇔\dfrac{1}{10-x²}=\dfrac{16x²-24x+9}{9x²}$
$⇔9x²=(10-x²)(16x²-24x+9)$
$⇔9x²=160x²-240x+90-16x^{4}+24x³-9x²$
$⇔16x^{4}-24x³-142x²+240x-90=0$
$⇔8x^{4}-12x³-71x²+120x-45=0$
…