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1. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel :

\(P=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{2}+\frac{9}{a+b+c}=\frac{3}{2}+\frac{9}{\frac{3}{2}}=\frac{3}{2}+6=\frac{15}{2}\)

Dấu "=" xảy ra <=> a=b=c=1/2

2. Áp dụng bất đẳng thức AM-GM :

\(P=\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\)

\(\ge2\sqrt{4a\cdot\frac{1}{a}}+2\sqrt{4b\cdot\frac{1}{b}}+2\sqrt{4c\cdot\frac{1}{c}}-3\cdot\frac{3}{2}=4\cdot3-\frac{9}{2}=\frac{15}{2}\)

Dấu "=" xảy ra <=> a=b=c=1/2

đk: \(x\le-1;x\ge-\frac{1}{4}\)

Đặt\(\hept{\begin{cases}\sqrt{4x^2+5x+1}=a\left(a\ge0\right)\\2\sqrt{x^2-x+1}=b\left(b\ge0\right)\end{cases}\Rightarrow\hept{\begin{cases}4x^2+5x+1=a^2\\4x^2-4x+4=b^2\end{cases}}}\)\(\Rightarrow a^2-b^2=a-b\Rightarrow\left(a-b\right)\left(a+b-1\right)=0\)

Vì \(b=\sqrt{4x^2-4x+4}\Rightarrow b\ge\sqrt{3}\Rightarrow a+b-1>0\Rightarrow a-b=0\)

\(\Rightarrow9x-3=0\Rightarrow x=\frac{1}{3}\left(TM\right)\)

đk: \(\hept{\begin{cases}x\le-\sqrt{2}\\x\ge\frac{1+\sqrt{5}}{2}\end{cases}}\)

Ta thấy \(\left(3x^2-5x+1\right)-3\left(x^2-x-1\right)=-2\left(x-2\right)\)

\(x^2-2-\left(x^2-3x+4\right)=3\left(x-2\right)\)

pt tương đương: \(\sqrt{3x^2-5x+1}-\sqrt{3\left(x^2-x-1\right)}=\sqrt{x^2-2}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\frac{-2\left(x-2\right)}{\sqrt{3x^2-5x+1}-\sqrt{3\left(x^2-x-1\right)}}=\frac{3\left(x-2\right)}{\sqrt{x^2-2}-\sqrt{x^2-3x+4}}\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{2}{\sqrt{3x^2-5x+1}-\sqrt{3\left(x^2-x-1\right)}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\right]=0\)

\(\Leftrightarrow x=2\) (tm)  ( Vì \(\frac{2}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}>0\))

\(b,ĐK:x\ge-2\)

\(\sqrt{4x+8}+\frac{1}{3}\sqrt{9x+18}=3\sqrt{\frac{x+2}{4}}+\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}+\sqrt{x+2}=\frac{3}{2}\sqrt{x+2}+\sqrt{2}\)

\(\Leftrightarrow\frac{3}{2}\sqrt{x+2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}=\frac{2\sqrt{2}}{3}\)

\(\Leftrightarrow x+2=\frac{8}{9}\)

\(\Leftrightarrow x=-\frac{10}{9}\left(tm\right)\)

\(b,\sqrt{4x+8}+\frac{1}{3}\sqrt{9x+18}=3\sqrt{\frac{x+2}{4}}+\sqrt{2}\)

\(ĐXKĐ:x\ge-2\)

\(2\sqrt{x+2}+\sqrt{x+2}=3\sqrt{\frac{x+2}{4}}+\sqrt{2}\)

\(\frac{3}{2}\sqrt{x+2}=\sqrt{2}\)

\(\sqrt{x+2}=\frac{2\sqrt{2}}{3}\)

\(x+2=\frac{8}{9}\)

\(x=\frac{-10}{9}\left(TM\right)\)

\(T=\left(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+\frac{x-81}{\sqrt{x}+9}\right).\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

\(T=\left(\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}+\frac{\left(\sqrt{x}-9\right)\left(\sqrt{x}+9\right)}{\sqrt{x}+9}\right).\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

\(T=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\sqrt{x}-9\right).\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

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

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

\(T=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right).\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

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

\(T=x-\sqrt{x}-6+\sqrt{x}\)

\(T=x-6\)

\(T=\left[\frac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}+\frac{\left(\sqrt{x}-9\right)\left(\sqrt{x}+9\right)}{\sqrt{x}+9}\right].\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

\(T=\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\sqrt{x}-9\right].\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

\(T=\left[x-\sqrt{x}+\sqrt{x}-9\right].\frac{\sqrt{x}+2}{\sqrt{x}+3}+\sqrt{x}\)

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

\(T=x-\sqrt{x}+6\)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}=x-\sqrt{x}-1\)

Ta có: \(\left[\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}\right]^2=3+\sqrt{2}+3-\sqrt{2}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\)

\(=6-2\sqrt{7}\)

\(\Leftrightarrow\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}=\sqrt{6-2\sqrt{7}}\)

\(\Leftrightarrow x=\frac{\sqrt{3-\sqrt{2}}+\sqrt{6-2\sqrt{7}}}{\sqrt{3+\sqrt{2}}}=1\)

\(P=x-\sqrt{x}-1=1-\sqrt{1}-1=-1\)

\(\sqrt{\frac{1}{4}x^2+\frac{1}{4}+1}=\sqrt{6-2\sqrt{5}}\)

\(\frac{1}{4}x^2+\frac{1}{4}+1=6-2\sqrt{5}\)

\(x^2=19-8\sqrt{5}\)

\(\orbr{\begin{cases}x=\sqrt{19-8\sqrt{5}}\\x=-\sqrt{19-8\sqrt{5}}\end{cases}}\)

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)

\(\Rightarrow A=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)