A B C H K E

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

B A C H E K E'

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

a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)

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

a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)

\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)

b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)

Từ giả thiết  => \(\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\le1-\frac{a}{a+1}=\frac{1}{a+1}\)

Áp dụng bđt Cauchy cho 3 số dương : \(\frac{1}{a+1}\ge\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3.\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\). Tương tự: \(\frac{1}{b+1}\ge3.\sqrt[3]{\frac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}}\)

\(\frac{1}{c+1}\ge3.\sqrt[3]{\frac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}}\)

\(\frac{1}{d+1}\ge3.\sqrt[3]{\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Nhân từ 4 bđt: \(1\ge81abcd\Rightarrow abcd\le\frac{1}{81}\)

a) đk: \(x\ge3\)

Ta có: \(\sqrt{x-3}=3x-11\)

\(\Leftrightarrow x-3=9x^2-66x+121\)

\(\Leftrightarrow9x^2-67x+124=0\)

\(\Leftrightarrow\left(9x^2-36x\right)-\left(31x-124\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(9x-31\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\9x-31=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{31}{9}\end{cases}}\)

a, \(\sqrt{x-3}=3x-11\left(đk:x\ge3\right)< =>\sqrt{x-3}-1=3x-12\)

\(< =>\frac{x-4}{\sqrt{x-3}+1}-3\left(x-4\right)=0< =>\left(x-4\right)\left(\frac{1}{\sqrt{x-3}+1}-3\right)=0\)

\(< =>\orbr{\begin{cases}x-4=0\\\frac{1}{\sqrt{x-3}+1}=3\end{cases}}< =>\orbr{\begin{cases}x=4\left(tm\right)\\\sqrt{x-3}+1=\frac{1}{3}\left(vl\right)\end{cases}}\)