ĐKXĐ: $-1≤x≤1$

Đặt $\sqrt[]{1-x}=a$;$\sqrt[]{1+x}=b$

$⇒a^2=1-x;b^2=1+x$

$⇒a^2+b^2=2$

và $x=b^2-1$

Khi đó pt trở thành:

$b^2-1+a^2+b^2=3ab+b$

$⇔2b^2+a^2-1=3ab+b$

$⇔2b^2-2ab-2b-ab+a^2+a+b-a-1=0$

$⇔2b(b-a-1)-a(b-a-1)+(b-a-1)=0$

$⇔(b-a-1)(2b-a+1)=0$

Nếu $b-a-1=0$

$⇔b=a+1$
$⇒\sqrt[]{1+x}=\sqrt[]{1-x}+1$

$⇔1+x=1-x+1+2.\sqrt[]{1-x}$

$⇔2x-1=2.\sqrt[]{1-x}$

$⇔\begin{cases}x≥\dfrac{1}{2}\\4x^2-4x+1=4-4x\end{cases}$

$⇔\begin{cases}x≥\dfrac{1}{2}\\4x^2-3=0\end{cases}$

$⇔\begin{cases}x≥\dfrac{1}{2}\\(\left[ \begin{array}{l}x=\sqrt[]{3}/2\\x=-\sqrt[]{3}/2\end{array} \right.\ \end{cases}$

$⇒x=\sqrt[]{3}/2$

Nếu $2b-a+1=0$

$⇔2b=a+1$

$⇔2.\sqrt[]{1+x}=\sqrt[]{1-x}+1$

$⇔4.(1+x)=1-x+1+2.\sqrt[]{1-x}$

$⇔5x+2=2.\sqrt[]{1-x}$

$⇔\begin{cases}x≥\dfrac{-2}{5}\\25x^2+20x+4=4-4x\end{cases}$

$⇔\begin{cases}x≥\dfrac{-2}{5}\\25x^2+24x=0\end{cases}$

$⇔\begin{cases}x≥\dfrac{-2}{5}\\(\left[ \begin{array}{l}x=0\\x=-\dfrac{24}{25}\end{array} \right.\ \end{cases}$

$⇔x=0(t/m)$

Vậy....$x=\sqrt[]{3}/2;0$

Đáp án:

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

Giải thích các bước giải:

$x+2=3\sqrt{(1-x)(1+x)}+\sqrt{1+x}$ $(*)$

$\text{ĐKXĐ: $-1 \leq x \leq 1$}$

$(*) ⇒ x^2+4x+4=9(1-x)(1+x)+1+x+6\sqrt{(1-x)(1+x)^2}$

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

$⇔ 10x^2+3x-6=6(1+x)\sqrt{1-x}$

$⇒ 100x^4+9x^2+36+60x^3-120x^2-36x=36(1+x)^2(1-x)$

$⇔ 100x^4+60x^3-111x^2-36x+36=36(1-x^2)(1+x)$

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

$⇔ 100x^4+96x^3-75x^2-72x=0$

$⇔ x(100x^3+96x^2-75x-72)=0$

$⇔ x(25x+24)(4x^2-3)=0$

$⇔ x(25x+24)(2x-\sqrt{3})(2x+\sqrt{3})=0$

$⇒ \left[ \begin{array}{l}x=0\\x=-\dfrac{24}{25}\\x=\dfrac{\sqrt{3}}{2}\\x=-\dfrac{\sqrt{3}}{2}\end{array} \right.$

$\text{Thử lại ta thấy $x_{1}=-\dfrac{24}{25}$; $x_{2}=\dfrac{\sqrt{3}}{2}$ là nghiệm phương trình.}$

$\text{Vậy phương trình có 2 nghiệm: $x_{1}=-\dfrac{24}{25}$; $x_{2}=\dfrac{\sqrt{3}}{2}$}$