\(ĐKXĐ:2x^2+16x+18\ge0;x^2-1\ge0\)

\(pt\Leftrightarrow\sqrt{x^2-1}=2x+4-\sqrt{2x^2+16x+18}\)(1)

\(\Leftrightarrow\sqrt{x^2-1}\left(\frac{2\sqrt{x^2-1}}{2x+4+\sqrt{2x^2+16x+18}}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-1}=0\\2\sqrt{x^2-1}=2x+4+\sqrt{2x^2+16x+18}\left(2\right)\end{cases}}\)

Lấy(1) + (2), ta được: \(3\sqrt{x^2-1}=4x+8\Leftrightarrow x=\frac{3\sqrt{57}-32}{7}\)

\(1\le a\le2\Rightarrow\left(a-1\right)\left(a-2\right)\le0\) \(\Rightarrow a^2-3a+2\le0\Rightarrow a^2+2\le3a\)

\(\Rightarrow a+\frac{2}{a}\le3\)\(\Rightarrow\left(a+\frac{2}{a}\right)^2\le9\Rightarrow a^2+\frac{4}{a^2}\le5\)

Tương tự : \(b+\frac{2}{b}\le3\); \(b^2+\frac{4}{b^2}\le5\)

\(\Rightarrow a+\frac{2}{a}+a^2+\frac{4}{a^2}+b+\frac{2}{b}+b^2+\frac{4}{b^2}\le16\)

Áp dụng BĐT Cô-si,ta có : 

\(16=\left(a+b^2+\frac{4}{a^2}+\frac{2}{b}\right)+\left(b+a^2+\frac{4}{b^2}+\frac{2}{a}\right)\ge2\sqrt{\left(a+b^2+\frac{4}{a^2}+\frac{2}{b}\right)\left(b+a^2+\frac{4}{b^2}+\frac{2}{a}\right)}\)

\(\Leftrightarrow8\ge\sqrt{\left(a+b^2+\frac{4}{a^2}+\frac{2}{b}\right)\left(b+a^2+\frac{4}{b^2}+\frac{2}{a}\right)}\)

\(\Leftrightarrow A=\left(a+b^2+\frac{4}{a^2}+\frac{2}{b}\right)\left(b+a^2+\frac{4}{b^2}+\frac{2}{a}\right)\le64\)

Vậy GTLN của A là 64 \(\Leftrightarrow\orbr{\begin{cases}a=b=1\\a=b=2\end{cases}}\)

\(pt\)\(\Leftrightarrow\)\(9 . ( x - 2 ) - ( x^2 - 4 )= 0\) ( bình phương vế lên )

\(\Leftrightarrow\)\(9. ( x - 2 ) - ( x + 2 )(x-2)=0\)

\(\Leftrightarrow\)\(( x - 2 )(7 - x )=0\)

\(\Leftrightarrow\)\(x - 2 = 0\) \(hoặc \)  \(7 - x = 0\)

\(\Leftrightarrow\)\(x = 2 \) \(hoặc\)  \(x= 7\)

\(ĐKXĐ:\hept{\begin{cases}x\ne9\\x\ne64\end{cases}}\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x-3}}+\frac{2\sqrt{x}-24}{x-9}\right).\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\left(\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x-3}\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)

\(\Leftrightarrow P=\frac{7}{\sqrt{x}+3}\)

Để P nguyên \(\Leftrightarrow7⋮\sqrt{x}+3\)    \(\left(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\right)\)

\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)
Ta có bảng sau :

Vậy \(x=16\Leftrightarrow P\in Z\)

Áp dụng BĐT Cô - si cho 2 số không âm, ta có:

\(VT=\text{Σ}_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\text{Σ}_{cyc}\sqrt{\frac{bc}{a}}\right)\)

\(\Leftrightarrow\text{Σ}_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)\)

\(+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)\)

\(\Leftrightarrow\text{Σ}_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)

\(+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

(Dấu "="\(\Leftrightarrow a=b=c=1\))

\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}=2\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)\)

\(=\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)+\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)\)

\(\ge2\sqrt{\sqrt{\frac{bc}{a}}\sqrt{\frac{ca}{b}}}+2\sqrt{\sqrt{\frac{ca}{b}}\sqrt{\frac{ab}{c}}}+2\sqrt{\sqrt{\frac{ab}{c}}\sqrt{\frac{bc}{a}}}\)

\(=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{a}\sqrt{b}\sqrt{c}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

Ta có BĐT cần chứng minh tương đương với:

\(\frac{a}{2}-\frac{a^2}{2a+1}+\frac{b}{2}-\frac{b^2}{2b+1}+\frac{c}{2}-\frac{c^2}{2c+1}\ge\frac{a+b+c}{2}-\frac{a^2+b^2+c^2}{\sqrt{a^2+b^2+c^2+6}}\)

Hay: \(\frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}+\frac{2\left(a^2+b^2+c^2\right)}{\sqrt{a^2+b^2+c^2+6}}\ge3\)

Áp dụng BĐT Bunhiacopxki dạng dạng p.thức ta được:

\(\frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}\ge\frac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2\right)+3}\)

Khi đó ta cần chứng minh:

\(\frac{9}{2\left(a^2+b^2+c^2\right)+3}+\frac{2\left(a^2+b^2+c^2\right)}{\sqrt{a^2+b^2+c^2+6}}\ge3\)

Đặt: \(t=a^2+b^2+c^2\ge3\) ta có:

\(\frac{9}{2t+3}+\frac{2t}{\sqrt{t+6}}\ge3\Leftrightarrow\frac{9}{2t+3}-1+\frac{2t}{\sqrt{t+6}}-2\ge0\)

\(\Leftrightarrow\frac{2\left(3-t\right)}{2t+3}+\frac{2t-2\sqrt{t+6}}{\sqrt{t+6}}\ge0\)

\(\Leftrightarrow\left(t-3\right)\left[\frac{t+2}{\sqrt{t+6}\left(t+\sqrt{t+6}\right)}-\frac{1}{2t+3}\right]\ge0\)

\(\Leftrightarrow\left(t+2\right)\left(2t+3\right)-\sqrt{t+6}\left(t+\sqrt{t+6}\right)\ge0\)

\(\Leftrightarrow t\left(2t+6-\sqrt{t+6}\right)\ge0\)

Vì: \(t\ge3\) nên BĐT luôn đúng.

BĐT xảy ra \(\Leftrightarrow a=b=c=1\)

Sử dụng Bunhiacopxki:

\(\sqrt{\left(\Sigma_{cyc}\frac{a^2}{\sqrt{a^2+b^2+c^2+6}}\right)\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\ge\Sigma_{cyc}\frac{a^2}{2a+1}=VT\)

Hay: \(\sqrt{VP.\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\ge VT\)

Vậy ta chỉ cần chứng minh: \(VP\ge\sqrt{VP.\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\)

\(\Leftrightarrow VP\ge\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\)

\(\Leftrightarrow\frac{a^2+b^2+c^2}{a^2+b^2+c^2+6}\ge\Sigma_{cyc}\frac{a^2}{\left(2a+1\right)^2}\)