a) Gọi AD là đường kính của ( O ; R ) 

Có: \(\Delta ADC\) nội tiếp đường tròn ( O ; R ) có O là trung điểm của AD \(\Rightarrow\)\(\Delta ADC\) vuông tại C 

Xét 2 tam giác vuông ABH và ADC có: ^ABH = ^ADC ( cùng chắn cung AC ) \(\Rightarrow\)\(\Delta ABH~\Delta ADC\) ( g - g ) 

\(\Rightarrow\)\(\frac{AB}{AD}=\frac{AH}{AC}\) hay \(\frac{c}{2R}=\frac{h}{b}\)\(\Leftrightarrow\)\(bc=2Rh\)

b) từ a ta có: \(bc=2Rh\)\(\Leftrightarrow\)\(\frac{abc}{4R}=\frac{2Rhc}{4R}=\frac{1}{2}hc=S_{ABC}\) ( đpcm ) 

a³ + b³ + c³ = a² + b² + c² = 1

\(\Rightarrow\)a² ( 1 - a ) + b² ( 1 - b ) + c² ( 1 - c ) = 0 ( 1 )

Mà a² + b² + c² = 1 \(\Rightarrow\)| a | \(\le\)1, | b | \(\le\)1 , | c | \(\le\)1

\(\Rightarrow\hept{\begin{cases}1-a\ge0\\1-b\ge0\\1-c\ge0\end{cases}}\Rightarrow\hept{\begin{cases}a^2\left(1-a\right)\ge0\\b^2\left(1-b\right)\ge0\\c^2\left(1-c\right)\ge0\end{cases}}\)

\(\Rightarrow\)a² ( 1 - a ) + b² ( 1 - b ) + c² ( 1 - c ) \(\ge\)0 ( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\hept{\begin{cases}a^2\left(1-a\right)=0\\b^2\left(1-b\right)=0\\c^2\left(1-c\right)=0\end{cases}}\)

( a,b,c ) là hoán vị của ( 0 ; 0 ; 1 )

Vậy S = 1

đề hơi sai

\(a,đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

\(b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow x=\sqrt{3}-1\)

\(\Rightarrow A=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}\)

\(b,A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}\)\(=1-\frac{4}{\sqrt{x}+2}\)

\(A\in Z\Leftrightarrow1-\frac{4}{\sqrt{x}+2}\in Z\Rightarrow\frac{4}{\sqrt{x}+2}\in Z\)

\(\Rightarrow\sqrt{x}+2\inƯ_4\)

Mà \(Ư_4=\left\{\pm1;\pm2;\pm4\right\}\)Nhưng \(\sqrt{x}+2\ge2\)\(\Rightarrow\sqrt{x}+2\in\left\{2;4\right\}\)

\(Th1:\sqrt{x}+2=2\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(Th2:\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(KL:x\in\left\{0;4\right\}\)

Có ai biết làm câu b Ko giúp t với 

chịu thôi

Áp dụng BĐT bunniacoxki ta có:

\(\left(b^2+\left(c+a\right)^2\right)\left(1+4\right)\ge\left(b+2\left(a+c\right)\right)^2\)

=> \(\sqrt{\frac{a^2}{b^2+\left(c+a\right)^2}}\le\sqrt{5}.\frac{a}{b+2c+2a}\)

=> \(VT\le\sqrt{5}.\left(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\right)\)

Cần CM \(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\le\frac{3}{5}\)

<=>\(\left(\frac{1}{2}-\frac{a}{b+2c+2a}\right)+\left(\frac{1}{2}-\frac{b}{c+2a+2b}\right)+\left(\frac{1}{2}-\frac{c}{a+2b+2c}\right)\ge\frac{9}{10}\)

<=>\(\frac{b+2c}{b+2c+2a}+\frac{c+2a}{c+2a+2b}+\frac{a+2b}{a+2b+2c}\ge\frac{9}{5}\)

Áp dụng bđt buniacoxki dạng phân thức ở vế trái:

=> \(VT\ge\frac{\left(b+2c+c+2a+a+2b\right)^2}{\left(b+2c\right)^2+2a\left(b+2c\right)+\left(c+2a\right)^2+2b\left(c+2a\right)+\left(a+2b\right)^2+2c\left(a+2b\right)}\)

         \(=\frac{9\left(a+b+c\right)^2}{5\left(a+b+c\right)^2}=\frac{9}{5}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

<=>\(9\sqrt{x}=6\)

<=>\(\sqrt{x}=\frac{6}{9}\)

<=>\(x=\left(\frac{6}{9}\right)^2=\frac{36}{81}=\frac{4}{9}\)