\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)

\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)

\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)

\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)

\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)

\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)

\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)

\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)

b. ĐKXĐ: \(x\ge5\)

\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)

\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)

\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2+3b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+5x+12}=a>0\\\sqrt{2x^2+3x+2}=b>0\end{matrix}\right.\) \(\Rightarrow x+5=\dfrac{a^2-b^2}{2}\)

Phương trình trở thành:

\(a+b=\dfrac{a^2-b^2}{2}\)

\(\Leftrightarrow\left(a-b-2\right)\left(a+b\right)=0\)

\(\Leftrightarrow a-b-2=0\) (do \(a+b>0\))

\(\Leftrightarrow a=b+2\)

\(\Leftrightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2\)

\(\Leftrightarrow2x^2+5x+12=2x^2+3x+6+4\sqrt{2x^2+3x+2}\)

\(\Leftrightarrow x+3=2\sqrt{2x^2+3x+2}\) (\(x\ge-3\))

\(\Leftrightarrow x^2+6x+9=4\left(2x^2+3x+2\right)\)

\(\Leftrightarrow7x^2+6x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{7}\end{matrix}\right.\)