Xét tam giác ABC vuông tại A ta có:

\(AB^2=BC\cdot BH\)

\(\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{\left(\dfrac{2}{3}\right)^2}{12}=\dfrac{1}{27}\left(cm\right)\)  

Mà: \(BC=CH+BH\)

\(\Rightarrow CH=12-\dfrac{1}{27}=\dfrac{323}{27}\left(cm\right)\)  

\(AC^2=BC\cdot CH\)

\(\Rightarrow AC=\sqrt{BC\cdot CH}=\sqrt{12\cdot\dfrac{323}{27}}=\dfrac{2\sqrt{323}}{3}\left(cm\right)\) 

Mà: \(AH\cdot BC=AB\cdot AC\)

\(\Rightarrow AH=\dfrac{AB\cdot AC}{BC}=\dfrac{\dfrac{2}{3}\cdot\dfrac{2\sqrt{323}}{3}}{12}=\dfrac{\sqrt{323}}{27}\left(cm\right)\)

\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

Để phương trình có nghiệm \(\Delta'\ge0\)

 \(\Rightarrow\left(\dfrac{10}{2}\right)^2-1.\left(2m+7\right)\ge0\\ 25-2m-7\ge0\\ \Leftrightarrow18-2m\ge0\\ \Leftrightarrow18\ge2m\\ \Leftrightarrow m\le9\)

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