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

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

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

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

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

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

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

Trước tiên ta chứng minh : \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\); ở đó, a,b tùy ý. Thật vậy:

\(2\left(a^2+b^2\right)S\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)

Ta có: \(x\left(x-1\right)+\frac{1}{4}+y\left(y-1\right)+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\ge\frac{1}{2}\text{[}\left(x-\frac{1}{2}\right)+\left(y-\frac{1}{2}\right)\text{]}^2\)\(=\frac{1}{2}\left(x+y-1\right)^2\ge\frac{1}{2}\left(6-1\right)^2=\frac{25}{2}\Rightarrow x\left(x-1\right)+y\left(y-1\right)\ge\frac{25}{2}-\frac{1}{2}=12\)

Khi x=y=3 thì => đpcm

Hơi khác cách của mình nhưng đúng r.

Ta có: 

\(2x=3y=6z\)

\(\Rightarrow\)\(\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}=\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=9\\\frac{y}{3}=9\\\frac{2z}{2}=9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=27\\y=18\\z=9\end{cases}}\)

xin cho tui sửa lại tí @@

Ta có: \(2x=3y=6z\)

\(=>\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(=>\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(=>\hept{\begin{cases}\frac{x}{3}=9=>x=9\cdot3=27\\\frac{y}{2}=9=>y=9\cdot2=18\\\frac{2z}{2}=9=>z=9\cdot2:2=9\end{cases}}\)

Vậy: x = 27

        y = 18

        z = 9

Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\)

Mặt khác : \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)\(\Rightarrow\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\ge\frac{3}{2}\)

Dự đoán \(MinL=\frac{3}{2}\)khi a = b = c

Ta cần chứng minh \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\Leftrightarrow\left(\frac{a}{a+b}-\frac{1}{2}\right)+\left(\frac{b}{b+c}-\frac{1}{2}\right)+\left(\frac{c}{c+a}-\frac{1}{2}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{\left(a-b\right)+\left(c-a\right)}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{a-b}{2\left(b+c\right)}-\frac{c-a}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2}\left(\frac{1}{a+b}-\frac{1}{b+c}\right)-\frac{c-a}{2}\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2}.\frac{c-a}{\left(a+b\right)\left(b+c\right)}-\frac{c-a}{2}.\frac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{\left(a-b\right)\left(c-a\right)\left(a+b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đúng do \(a\ge b\ge c>0\))

Đẳng thức xảy ra khi a = b = c