Question

Доказать справедливость равенства​

Answer 1

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

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

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

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

$\left(\frac{4x^4+6x^2-2x^3\sqrt{4x^2+3}+2x^2+3-x\sqrt{4x^2+3}+2x^3\sqrt{4x^2+3}+3x\sqrt{4x^2+3}-x^2(4x^2+3)}{(2x^2+3)^2-(x\sqrt{4x^2+3})^2}\right)^{\frac{1}{2}}=$

$\left(\frac{4x^4+6x^2-2x^3\sqrt{4x^2+3}+2x^2+3-x\sqrt{4x^2+3}+2x^3\sqrt{4x^2+3}+3x\sqrt{4x^2+3}-4x^4-3x^2}{4x^4+12x^2+9-4x^4-3x^2}\right)^{\frac{1}{2}}=$

$\left(\frac{5x^2+2x\sqrt{4x^2+3}+3}{9x^2+9}\right)^{\frac{1}{2}}=$

$\left(\frac{x^2+2x\sqrt{4x^2+3}+4x^2+3}{9(x^2+1)}\right)^{\frac{1}{2}}=$

$\left(\frac{x^2+2x\sqrt{4x^2+3}+(\sqrt{4x^2+3})^2}{9(x^2+1)}\right)^{\frac{1}{2}}=$

$\left(\frac{(x+\sqrt{4x^2+3})^2}{9(x^2+1)}\right)^{\frac{1}{2}}=$

$\frac{x+\sqrt{4x^2+3}}{3\sqrt{x^2+1}}$