Đáp án:

Giải thích các bước giải: Tham khảo

ĐKXĐ $: - 1 ≤ x ≤ 1$

Đặt $ : u = \sqrt{1 + x} ≥ 0; v = \sqrt{1 - x} ≥ 0$

$⇒ u² + v² = 2 ⇔ v² = 2 - u² (1)$

$ PT ⇔ u² + 1 = 3uv + u ⇔ 3uv = u² - u + 1$

$ ⇔ 9u²v² = u^{4} + u² + 1 - 2u³ - 2u + 2u²$

$ ⇔ 9u²(2 - u²) = u^{4} - 2u³ + 3u² - 2u + 1$

$ ⇔ 10u^{4} - 2u³ - 15u² - 2u + 1 = 0$

$ ⇔ (u + 1)(5u - 1)(2u² - 2u - 1) = 0$

@ $ 5u - 1 = 0 ⇔ u = \dfrac{1}{5} ⇒ x = - \dfrac{24}{25}$ 

@ $ 2u² - 2u - 1 = 0$

$ ⇒  u = \dfrac{1 + \sqrt{3}}{2} ⇒ x = \dfrac{\sqrt{3}}{2}$ ( loại $: u = \dfrac{1 - \sqrt{3}}{2} < 0)$

$x+2=3\sqrt{1-x^2}+\sqrt{1+x}_{}$ ĐK: $1_{}$ $\geq$ $x_{}$ $\geq-1$ 

$⇔x^2+4x+4=9.(1-x^2)+6\sqrt{(1-x^2)(1+x)}+1+x_{}$

$⇔x^2+4x+4=9-9x^2+6\sqrt{1+x-x^2-x^3}+1+x_{}$

$⇔x^2+4x+4=10-9x^2+6\sqrt{1+x-x^2-x^3}+x_{}$

$⇔-6\sqrt{1+x-x^2-x^3}=10-9x^2+x-x^2-4x-4_{}$

$⇔-6\sqrt{1+x-x^2-x^3}=6-10x^2-3x_{}$

$⇔36.(1+x-x^2-x^3)=36+100x^4+9x^2-120x^2-36x+60x^3_{}$

$⇔36+36x-36x^2-36x^3=36+100x^4+9x^2-120x^2-36x+60x^3_{}$

$⇔36x-36x^2-36x^3=100x^4+9x^2-120x^2-36x+60x^3_{}$

$⇔36x-36x^2-36x^3=100x^4-111x^2-36x+60x^3_{}$

$⇔36x-36x^2-36x^3-100x^4+111x^2+36x-60x^3=0_{}$

$⇔72x+75x^2-96x^3-100x^4=0_{}$

$⇔x.(72+75x-96x^2-100x^3)=0_{}$

$⇔x.[ 3.(24+25x)-4x^2.(24+25x)]=0 _{}$

$⇔x.(24+25x)(3-4x^2)=0_{}$

$*)_{}$ $th1:_{}$ $x=0_{}$ $(KTMĐK)_{}$ 

$*)_{}$ $th2:_{}$ $24+25x=0_{}$

$⇔x=_{}$ $-\frac{24}{25}$ $(TMĐK)_{}$ 

$*)_{}$ $th3:_{}$ $3-4x^2=0_{}$

$⇔(\sqrt{3}-2x)(\sqrt{3}+2)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}\sqrt{3}-2x=0\\\sqrt{3}+2x=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=\frac{\sqrt{3}}{2}(TMĐK)\\x=-\frac{\sqrt{3}}{2}(KTMĐ)\end{array} \right.\)

$Vậy_{}$ $x_{1}=$ $-\frac{24}{25};$ $x_{2}=$ $\frac{\sqrt{3}}{2}$

$*_{}$ $nguyencuong18082006_{}$

$*_{}$ $Simple_{}$ $Love_{}$