ĐKXĐ:$x\geq \dfrac{-7}{8}$
[tex]a)\sqrt{4x^{2}+6}-\sqrt{8x+7}=8x^{3}+40x^{2}-8x-29\\\Leftrightarrow \sqrt{4x^{2}+6}-\sqrt{8x+7}+2x-1=8x^{3}+40x^{2}-8x-29+2x-1\\\Leftrightarrow (\sqrt{4x^{2}+6}-3)+[(2x+2)-\sqrt{8x+7}]=8x^3+40x^2-6x-30(1)[/tex]
Ta có:
[tex](\sqrt{4x^2+6}-3)(\sqrt{4x^2+6}+3)=4x^2+6-9=4x^2-3\\\Rightarrow \sqrt{4x^2+6}-3=\dfrac{4x^2-3}{\sqrt{4x^2+6}+3}[/tex]
[tex][(2x+2)-\sqrt{8x+7}][(2x+2)+\sqrt{8x+7}]\\=(2x+2)^2-(8x+7)\\=4x^2+8x+4-8x-7\\=4x^2-3\\\Rightarrow (2x+2)-\sqrt{8x+7}=\dfrac{4x^2-3}{(2x+2)+\sqrt{8x+7}}[/tex]
[tex]8x^3+40x^2-6x-30\\=2(4x^3+20x^2-3x-15)\\=2[4x^2(x+5)-3(x+5)]\\=2(x+5)(4x^2-3)[/tex]
Khi đó pt $(1)$ trở thành
$\dfrac{4x^2-3}{\sqrt{4x^2+6}+3}+\dfrac{4x^2-3}{(2x+2)+\sqrt{8x+7}}=2(x+5)(4x^2-3)$
[tex]\dfrac{4x^2-3}{\sqrt{4x^2+6}+3}+\dfrac{4x^2-3}{(2x+2)+\sqrt{8x+7}}=2(x+5)(4x^2-3)\\\Leftrightarrow \dfrac{4x^2-3}{\sqrt{4x^2+6}+3}+\dfrac{4x^2-3}{(2x+2)+\sqrt{8x+7}}-2(x+5)(4x^2-3)=0\\\Leftrightarrow (4x^2-3)\left (\dfrac{1}{\sqrt{4x^2+6}+3}+\dfrac{1}{2x+2+\sqrt{8x+7}}-2x-10 \right )=0\\\\..........\\\Leftrightarrow x=\pm \dfrac{\sqrt{3}}{2}[/tex]