a)ĐKXĐ:`x≥-10/3`
`x^2+9x+20=2\sqrt{3x+10}`
`⇔x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0`
`⇔(x+3)^2+(\sqrt{3x+10)-1)^2=0`
`⇔`$\left \{ {{x+3=0} \atop {\sqrt{3x+10}-1=0}} \right.$
`⇔`$\left \{ {{x=-3} \atop {3x+10=1}} \right.$
`⇔`$\left \{ {{x=-3} \atop {x=-3}} \right.$
Vậy `x=-3`
b)ĐKXĐ: `x≥0`
`8x^2-3x+6=4x\sqrt{3x^2+x+2}`
`⇔3x^2+x+2-4\sqrt{3x^2+x+2}+4x^2+x^2-4x+4=0`
`⇔(\sqrt{3x^2+x+2}-2x)^2+(x-2)^2=0`
`⇔`$\left \{ {{\sqrt{3x^2+x+2}-2x=0} \atop {x-2=0}} \right.$
`⇔`$\left \{ {{3x^2+x+2=4x^2} \atop {x=2}} \right.$
`⇔`$\left \{ {{x^2-x-2=0} \atop {x=2}} \right.$
`⇔`$\left \{ {{(x-2)(x+1)=0} \atop {x=2}} \right.$
`⇔`$\left \{ {{ \left[ \begin{array}{l}x=2\\x=-1(KTM)\end{array} \right. } \atop {x=2}} \right.$
Vậy `x=2`
