ĐK: \(x\ge4\).

\(A=\frac{1}{x-4\sqrt{x-4}+3}=\frac{1}{x-4-4\sqrt{x-4}+4+3}=\frac{1}{\left(\sqrt{x-4}-2\right)^2+3}\le\frac{1}{3}\)

Dấu \(=\)khi \(\sqrt{x-4}-2=0\Leftrightarrow x=8\)(thỏa mãn) 

ĐK: \(x\ge0,x\ne1\).

a) Khi \(x=9\): 

\(A=\frac{4\sqrt{x}+6}{x-1}=\frac{4.\sqrt{9}+6}{9-1}=\frac{18}{8}=\frac{9}{4}\).

b) \(B=\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{x+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

\(M=\frac{A}{B}=\frac{4\sqrt{x}+6}{x-1}\left(\sqrt{x}-1\right)=\frac{4\sqrt{x}+6}{\sqrt{x}+1}\)

c) \(M=\frac{4\sqrt{x}+6}{\sqrt{x}+1}=\frac{4\sqrt{x}+4+2}{\sqrt{x}+1}=4+\frac{2}{\sqrt{x}+1}\le4+\frac{2}{0+1}=6\)

Dấu \(=\)khi \(x=0\).

a, BC=BH+HC=8BC=BH+HC=8

Áp dụng HTL: 

⎧⎪⎨⎪⎩AB2=BH⋅BC=16AC2=CH⋅BC=48AH2=CH⋅BC=12⇒⎧⎪ ⎪⎨⎪ ⎪⎩AB=4(cm)AC=4√3(cm)AH=2√3(cm){AB2=BH⋅BC=16AC2=CH⋅BC=48AH2=CH⋅BC=12⇒{AB=4(cm)AC=43(cm)AH=23(cm)

b,b, Vì K là trung điểm AC nên AK=12AC=2√3(cm)AK=12AC=23(cm)

Ta có tanˆAKB=ABAK=42√3=2√33≈tan490tan⁡AKB^=ABAK=423=233≈tan⁡490

⇒ˆAKB≈490