Question

Answer 1

b: ΔHAB vuông tại H có HM là trung tuyến

nên \(HM=\dfrac{AB}{2}\)

ΔHAC vuông tại H có HN là trung tuyến

nên \(HN=\dfrac{AC}{2}\)

\(\left(\dfrac{HA}{HM}\right)^2+\left(\dfrac{HA}{HN}\right)^2\)

\(=\left(HA:\dfrac{AB}{2}\right)^2+\left(HA:\dfrac{AC}{2}\right)^2\)

\(=4\cdot HA^2\left(\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\right)=4\cdot\dfrac{HA^2}{HA^2}=4\)

a: 

\(A=\sqrt{2}\cdot\dfrac{\sqrt{5-2\sqrt{6}}-1}{\sqrt{4-2\sqrt{3}}-1}\)

\(=\sqrt{2}\cdot\dfrac{\sqrt{3}-\sqrt{2}-1}{\sqrt{3}-1-1}=\dfrac{\sqrt{2}}{-2+\sqrt{3}}\cdot\left(\sqrt{3}-\sqrt{2}-1\right)\)

\(=\left(-\sqrt{3}+\sqrt{2}+1\right)\cdot\sqrt{2}\left(2+\sqrt{3}\right)\)