Đáp án: $ P=\dfrac{3+\sqrt{3}}{2}$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}$
$=\dfrac{2\sqrt{3}+2-3(2\sqrt{3}-2)}{(2\sqrt{3}-2)(2\sqrt{3}+2)}$
$=\dfrac{8-4\sqrt{3}}{8}$
$=\dfrac{4-2\sqrt{3}}{4}$
$=\dfrac{3-2\sqrt{3}+1}{4}$
$=\dfrac{(\sqrt{3}-1)^2}{4}$
$=(\dfrac{\sqrt{3}-1}{2})^2$
$\to x=\sqrt{(\dfrac{\sqrt{3}-1}{2})^2}$
$\to x=\dfrac{\sqrt{3}-1}{2}$
$\to 2x=\sqrt{3}-1$
$\to 2x+1=\sqrt3$
$\to (2x+1)^2=3$
$\to 4x^2+4x+1=3$
$\to 4x^2+4x-2=0$
$\to 2x^2+2x-1=0$
Ta có:
$P=\dfrac{4(x+1)x^{2020}-2x^{2019}+2x+1}{2x^2+3x}$
$\to P=\dfrac{2x^{2019}(2(x+1)x-1)+2x+1}{2x^2+3x}$
$\to P=\dfrac{2x^{2019}(2x^2+2x-1)+2x+1}{2x^2+2x-1+x+1}$
$\to P=\dfrac{2x^{2019}\cdot 0+2x+1}{0+x+1}$ vì $2x^2+2x-1=0$
$\to P=\dfrac{2x+1}{x+1}$
$\to P=\dfrac{x^2+2x-1-x^2+1}{x+1}$
$\to P=\dfrac{0-x^2+1}{x+1}$
$\to P=-\dfrac{x^2-1}{x+1}$
$\to P=-\dfrac{(x-1)(x+1)}{x+1}$
$\to P=-(x-1)$
$\to P=1-x$
$\to P=1-\dfrac{\sqrt{3}-1}{2}$
$\to P=\dfrac{3+\sqrt{3}}{2}$
