b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)

Xét \(\Delta ABC\)vuông tại A có: \(AH\perp BC\Rightarrow\hept{\begin{cases}AB^2=BH.BC\\AC^2=CH.BC\end{cases}}\)

Ta có: \(AB^2.HC=BH.BC.HC\left(1\right)\)

\(AC^2.HB=CH.BC.HC\left(2\right)\)

Từ 1 và 2= đpcm

 Xét hai tam giác vuông ABH và CAH có:

ABH^=HAC^ (cùng phụ với góc BAH^)

Do đó, ΔABH∼ΔCAH

Suy ra: AH/CH=BH/AH ⇒AH^2=BH.CH.         

\(B=\frac{\sqrt{3-2\sqrt{2}}-\sqrt{3}}{\sqrt{5-2\sqrt{6}}+1}=\frac{\sqrt{\left(\sqrt{2}\right)^2-2.1.\sqrt{2}+1^2}-\sqrt{3}}{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}+1}=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{3}}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+1}=\frac{\sqrt{2}-1-\sqrt{3}}{\sqrt{3}-\sqrt{2}+1}=-1\)

a) 

Ta có: \(\frac{OD}{AD}=\frac{S_{BOC}}{S_{ABC}};\frac{OE}{BE}=\frac{S_{AOC}}{S_{ABC}};\frac{OF}{CF}=\frac{S_{AOB}}{S_{ABC}}\)\(\Rightarrow\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}=\frac{S_{BOC}+S_{AOC}+S_{AOB}}{S_{ABC}}\)

\(\Rightarrow\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}=\frac{S_{ABC}}{S_{ABC}}=1\left(ĐPCM\right)\)

b) chịu

b) Gợi ý nhỏ: Min=64

Khá ez:))

Δ AKB ~ Δ AEC (g.g) vì:

+ \(\widehat{BAK}=\widehat{CAE}\) (góc chung)

+ \(\widehat{AKB}=\widehat{AEC}=90^0\)

=> \(\frac{AK}{AE}=\frac{AB}{AC}\)

Từ đó ta dễ dàng CM được: Δ AKE ~ Δ ABC (c.g.c)

=> \(\frac{S_{AKE}}{S_{ABC}}=\left(\frac{AK}{AB}\right)^2=\cos^2A\)

Tương tự như vậy ta CM được: \(\frac{S_{BHE}}{S_{ABC}}=\cos^2B\) ; \(\frac{S_{CHK}}{S_{ABC}}=\cos^2C\)

Thay vào ta sẽ được: \(\left(1-\cos^2A-\cos^2B-\cos^2C\right)\cdot S_{ABC}\)

\(=\left(1-\frac{S_{AKE}}{S_{ABC}}-\frac{S_{BHE}}{S_{ABC}}-\frac{S_{CHK}}{S_{ABC}}\right)\cdot S_{ABC}\)

\(=S_{ABC}-S_{AKE}-S_{BHE}-S_{CHK}=S_{HKE}\)

=> đpcm

Kẻ EE' vuông góc với AC., ta có:

\(\frac{S_{AKE}}{S_{ABC}}=\frac{\frac{1}{2}EE'.AK}{\frac{1}{2}BK.AC}=\frac{EE'}{BK}.\frac{AK}{AC}=\frac{AE}{AB}.\frac{AK}{AC}\)

\(=\frac{AE}{AC}.\frac{AK}{AB}=\cos A.\cos A=\cos^2A.\)

Vậy \(\frac{S_{AKE}}{S_{ABC}}=\cos^2A\)

Tương tự, \(\frac{S_{BEH}}{S_{ABC}}=\cos^2B;\frac{S_{CKH}}{S_{ABC}}=\cos^2C\)\(\Rightarrow\frac{S_{KHE}}{S_{ABC}}=1-\frac{S_{AKE}}{S_{ABC}}-\frac{S_{BEH}}{S_{ABC}}-\frac{S_{CKH}}{S_{ABC}}=1-\cos^2A-\cos^2B-\cos^2C\)

Vậy =>đpcm

ĐK: \(x\ge0\)

Với \(x\ge0\Rightarrow\sqrt{\left(x+1\right)^3}-\sqrt{x}>0\)nên bpt \(\Leftrightarrow\sqrt{x\left(x+2\right)}\ge\sqrt{\left(x+1\right)^3}-\sqrt{x}\)

\(\Leftrightarrow x^2+2x\ge x^3+3x^2+4x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\)

\(\Leftrightarrow x^3+2x^2+2x+1-2\left(x+1\right)\sqrt{x\left(x+1\right)}\le0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2+x+1-2\sqrt{x\left(x+1\right)}\right]\le0\)

\(\Leftrightarrow x^2+x+1-2\sqrt{x\left(x+1\right)}\le0\)

\(\Leftrightarrow\left(\sqrt{x\left(x+1\right)}-1\right)^2\le0\Leftrightarrow\sqrt{x\left(x+1\right)}-1=0\)

\(\Leftrightarrow x=\frac{-1\pm\sqrt{5}}{2}.dox\ge0\Rightarrow x=\frac{-1+\sqrt{5}}{2}\)