Sửa: \(\dfrac{1}{3}\left(1,5v_1-v_1\right)=5\Leftrightarrow v_1=30\left(km\text{/}h\right)\)

Thay vào: \(\Rightarrow v_2=30\cdot1,5=45\left(km\text{/}h\right)\)

Quãng đường AB là:

\(s_{AB}=v_1\cdot t_1=v_2\cdot t_2=3v_1=2v_2\left(km\right)\)

\(\Leftrightarrow1,5v_1=v_2\)

Vì sau 20 phút \(=\dfrac{1}{3}\) giờ, hai xe cách nhau 5 km:

\(\dfrac{1}{3}\left(v_2-v_1\right)=5\left(km\right)\)

\(\Leftrightarrow\dfrac{1}{3}\left(1,5v_1-v_1\right)=5\left(km\right)\)

\(\Leftrightarrow\dfrac{1}{3}\cdot0,5v_1=5\left(km\right)\)

\(\Leftrightarrow v_1=36\left(km\text{/}h\right)\)

Thay vào: \(\Rightarrow v_2=36\cdot1,5=54\left(km\text{/}h\right)\)

\(\sqrt{x^2-x}=\sqrt{3x-5}\) ĐK: \(x\ge\dfrac{5}{3}\)

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

\(\Leftrightarrow x^2-x=3x-5\)

\(\Leftrightarrow x^2-x-3x+5=0\)

\(\Leftrightarrow x^2-4x+5=0\)

⇒ Không tách được thành hạng tử.

Vậy pt vô nghiệm.

\(-\dfrac{5}{12}-3,7-\dfrac{7}{12}-6,3\)

\(=-\dfrac{5}{12}-\dfrac{37}{10}-\dfrac{7}{12}-\dfrac{63}{10}\)

\(=\left(-\dfrac{5}{12}-\dfrac{7}{12}\right)-\left(\dfrac{37}{10}+\dfrac{63}{10}\right)\)

\(=-\dfrac{12}{12}-\dfrac{100}{10}\)

\(=-1-10\)

\(=-11\)

\(\sqrt{x^2-x-6}=\sqrt{x-3}\) ĐK: \(x\ge3\)

\(\Leftrightarrow\sqrt{\left(x^2-x-6\right)^2}=\sqrt{\left(x-3\right)^2}\)

\(\Leftrightarrow x^2-x-6=x-3\) 

\(\Leftrightarrow x^2-x-x-6+3=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(\text{loại}\right)\\x=3\left(\text{TM}\right)\end{matrix}\right.\)

Vậy \(S=\left\{3\right\}\).

\(\sqrt{27+10\sqrt{2}}-3\sqrt{2}=\sqrt{\left(5+\sqrt{2}\right)^2}-3\sqrt{2}=5+\sqrt{2}-3\sqrt{2}=5-2\sqrt{2}\)