`a)`\(D=\dfrac{a+1}{\sqrt{a}}+\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}+\dfrac{a^2-a\sqrt{a}+\sqrt{a}-1}{\sqrt{a}-a\sqrt{a}}\);\(a\ge0\)

\(D=\dfrac{a+1}{\sqrt{a}}+\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)}{\sqrt{a}\left(a-1\right)}\)

\(D=\dfrac{a+1}{\sqrt{a}}+\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{\left(\sqrt{a}-1\right)\left(a\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(D=\dfrac{a+1+a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(D=\dfrac{2a+\sqrt{a}+2}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\)

\(D=\dfrac{2a+\sqrt{a}+2-a+\sqrt{a}-1}{\sqrt{a}}\)

\(D=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}}\)

\(D=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)

\(\dfrac{1}{5}\sqrt{25.2}-2\sqrt{16.6}-\sqrt{2}+\sqrt{\dfrac{144}{6}}=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)

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bổ sung H là trực tâm, cm HD = DK ( sửa đề ) 

Xét tứ giác CEFB có ^CEB = ^CFB = 90⁰

mà 2 góc này kề, cùng nhìn cạnh BC 

Vậy tg CEFB nt 1 đường tròn 

=> ^EFC = ^EBC ( 2 góc nt chắn cung EC ) 

Xét tứ giác AFHE có 

^AFH + ^AEH = 180⁰

mà 2 góc này đối 

Vậy tg AFHE nt 1 đường tròn 

=> ^HFE = ^HAE ( 2 góc nt chắn cung EH ) 

=> ^EAH = ^EBC 

mà ^CAK = ^CBK ( 2 góc nt chắn cung CK của đường tròn O ) 

=> ^HBD = ^DBK 

Vậy tam giác HBK cân tại B 

mà BD là đường cao đồng thời là đường trung tuyến 

=> HD = DK 

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Đặt \(P=\dfrac{a}{\sqrt{b^3+1}}+\dfrac{b}{\sqrt{c^3+1}}+\dfrac{c}{\sqrt{a^3+1}}\)

\(=\dfrac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\dfrac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\dfrac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)

\(\Rightarrow P\ge\dfrac{2a}{b+1+b^2-b+1}+\dfrac{2b}{c+1+c^2-c+1}+\dfrac{2c}{a+1+a^2-a+1}=\dfrac{2a}{b^2+2}+\dfrac{2b}{c^2+2}+\dfrac{2c}{a^2+2}\)

Mặt khác với mọi \(a>0\) ta có:

\(\dfrac{1}{a^2+2}\ge\dfrac{7-2a}{18}\)

Thật vậy, BĐT trên tương đương:

\(18-\left(7-2a\right)\left(a^2+2\right)\ge0\)

\(\Leftrightarrow\left(a-2\right)^2\left(2a+1\right)\ge0\) (luôn đúng)

Từ đó \(\Rightarrow P\ge\dfrac{2a\left(7-2b\right)}{18}+\dfrac{2b\left(7-2c\right)}{18}+\dfrac{2c\left(7-2a\right)}{18}\)

\(\Rightarrow P\ge\dfrac{7\left(a+b+c\right)}{9}-\dfrac{2\left(ab+bc+ca\right)}{9}\ge\dfrac{7\left(a+b+c\right)}{9}-\dfrac{2\left(a+b+c\right)^2}{27}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=2\)

a. ĐKXĐ: \(x\ge0;x\ne1\)

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

\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b.

\(A=\dfrac{2\sqrt{x}+2-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1>0;\forall x\Rightarrow\dfrac{3}{\sqrt{x}+1}>0\)

\(\Rightarrow2-\dfrac{3}{\sqrt{x}+1}< 2\)

Hay \(A< 2\)

Điều kiện : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

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

\(\Leftrightarrow A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow A=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b) Ta có : 

\(A-2=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-1-2\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+1}\)

Mặt khác : -3 < 0 và \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow A-2< 0\Leftrightarrow A< 2\)