Giải các phương trình sau:
$a)3\sqrt{x+2}-6\sqrt{2-x}+4\sqrt{4-x^2}=10-3x$
$b)\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0$
$c)\sqrt{2x-1}+x^2-3x+1=0$
$d)\sqrt{\frac{x-7}{8}}+\sqrt{\frac{x-6}{9}}+\sqrt{\frac{x-5}{10}}=\sqrt{\frac{x-8}{7}}+\sqrt{\frac{x-9}{6}}+\sqrt{\frac{x-10}{5}}$
$e)\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}$
$f)\sqrt[3]{x+1}+\sqrt[3]{x^2}=\sqrt[3]{x}+\sqrt[3]{x^2+x}$