Đáp án:
$A=3-\sqrt{3}$
Giải thích các bước giải:
$x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2(\sqrt{3}+1)}}$
$⇔ x^2=\dfrac{\sqrt{3}+1-3(\sqrt{3}-1)}{2(\sqrt{3}-1)(\sqrt{3}+1)}$
$⇔ x^2=\dfrac{4-2\sqrt{3}}{2(3-1)}$
$⇔ x^2=\dfrac{(\sqrt{3}-1)^2}{4}$
$⇒ x=\dfrac{\sqrt{3}-1}{2}$ (vì $x \geq 0$)
$⇒ 2x=\sqrt{3}-1$
Ta có: $A=\dfrac{4(x+1)x^{2019}-2x^{2018}+2x+1}{2x^2+3x}$
$=\dfrac{2x(2x+2).x^{2018}-2x^{2018}+2x+1}{x(2x+3)}$
$=\dfrac{(\sqrt{3}-1)(\sqrt{3}+1).x^{2018}-2x^{2018}+\sqrt{3}}{\dfrac{\sqrt{3}-1}{2}.(\sqrt{3}+2)}$
$=\dfrac{(3-1).x^{2018}-2.x^{2018}+\sqrt{3}}{\dfrac{1+\sqrt{3}}{2}}$
$=\dfrac{2\sqrt{3}}{1+\sqrt{3}}$
$=\dfrac{2\sqrt{3}(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$
$=\dfrac{2\sqrt{3}-6}{1-3}$
$=3-\sqrt{3}$
