Giải:
1; ĐKXĐ:$-1$ \leq $x$ \leq $1$
$4\sqrt{x+1}=2(x+1)-(1-x)+2\sqrt{1-x}+\sqrt{(1-x)(1+x)}$ Đặt $\sqrt{1-x}=a; \sqrt{1-x}=b$
$\iff 4b=2b^2-a^2+2a+ab$
$\iff (2b-a)(b+a)-2(2b-a)=0$
$\iff (2b-a)(b+a-2)=0$
$\iff 2b=a$ v $b+a=2$
2, ĐKXĐ: $4-x^2$ \geq $0$
$(x+\sqrt{4-x^2})^2=(2+3x\sqrt{4-x^2})^2$
$\iff 4+2x\sqrt{4-x^2}=(2+3x\sqrt{4-x^2})^2$ Đặt $\sqrt{4-x^2}=a (a$ \geq $0)$
$\iff 4+2a=4+12a+9a^2$
$\iff 9a^2+10a=0$
$\iff a=0$
$\iff 4=x^2$
$\iff x=2$ v $x=-2$