a: Thay x=16 vào B, ta được:

\(B=\dfrac{4+3}{4-2}=\dfrac{7}{2}\)

\(K=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=8-2\sqrt{15}\)

\(1,\Delta=b^2-4ac=5^2-4.2=17>0\)

=> Pt có 2n pb

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-5+\sqrt{17}}{4}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-\sqrt{17}}{4}\)

2,\(\Delta=b^2-4ac=\left(-1\right)^2-4.5.\left(-2\right)=41>0\)

=> Pt có 2n pb

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+\sqrt{41}}{10}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-\sqrt{41}}{10}\)

3,\(\Delta=b^2-4ac=\left(-1\right)^2-4.9=-35< 0\)

=> Pt vô nghiệm

a: Để (d)//y=x-3 thì \(4-m^2=1\)

\(\Leftrightarrow m\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

\(M=\sqrt{\dfrac{\left(11+\sqrt{96}\right)^2}{121-96}}+\sqrt{\dfrac{\left(11-\sqrt{96}\right)^2}{121-96}}\\ M=\dfrac{11+\sqrt{96}}{5}+\dfrac{11-\sqrt{96}}{5}=\dfrac{22}{5}\)

\(D=\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{2-\sqrt{3}}-\dfrac{6}{3-\sqrt{3}}\)

\(=\sqrt{3}+1-6-3\sqrt{3}-3-\sqrt{3}\)

\(=-8-3\sqrt{3}\)