a) \(x^2+9x+20=2\sqrt{3x+10}\)

\(\Leftrightarrow\left(x+4\right)^2\left(x+5\right)^2=4\left(3x+10\right)\)

\(\Leftrightarrow x^4+10x^3+25x^2+8x^3+80x^3+200x+16x^2+160x+400=12x+40\)

\(\Leftrightarrow x^4+18x^3+121x^2+360x+400=12x+40\)

\(\Leftrightarrow x^4+18x^2+121x^2+360x+400-12x-40=0\)

\(\Leftrightarrow\left(x^3+15x^2+76x+120\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2+12x+40\right)\left(x+3\right)=0\)

Nhưng \(x^2+12x+40\ne0\), nên:

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: nghiệm phương trình là {-3}

ĐKXĐ: \(x\ge-\dfrac{10}{3}\)

\(\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}-1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-3\)

ĐK:....

\(x^2+9x+20=2\sqrt{3x+10}\)

\(\Leftrightarrow x^2+9x+20-2\sqrt{3x+10}=0\)

\(\Leftrightarrow x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\)

\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3x+10=1\end{matrix}\right.\)

\(\Leftrightarrow x=-3\)

Vậy....

\(x\ge-\frac{10}{3}\)

\(x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\)

\(\Leftrightarrow\left(x+3\right)^2-\left(\sqrt{3x+10}-1\right)^2=0\)

\(\Leftrightarrow\left(x+4-\sqrt{3x+10}\right)\left(x+2+\sqrt{3x+10}\right)=0\)

TH1: \(x+4-\sqrt{3x+10}=0\Leftrightarrow\left\{{}\begin{matrix}x+4\ge0\\\left(x+4\right)^2=3x+10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x^2+5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

TH2: \(x+2+\sqrt{3x+10}=0\Leftrightarrow\left\{{}\begin{matrix}-x-2\ge0\\\left(-x-2\right)^2=3x+10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-3\end{matrix}\right.\)