\(Pt\Leftrightarrow\sqrt{\left(x-\frac{1}{2}\right)^2}=\left(2x-1\right)\left(x^2+1\right).\)

(Đk có nghiệm: \(x\ge\frac{1}{2}\))

\(Pt\Leftrightarrow\left|x-\frac{1}{2}\right|=\left(2x-1\right)\left(x^2+1\right)\Rightarrow x-\frac{1}{2}=\left(2x-1\right)\left(x^2+1\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2+1-\frac{1}{2}\right)=0\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\left(t.m\right)\)

Áp dụng BĐT Cô - Si cho hai số dương \(ab\)và \(\frac{1}{ab}\), ta có : 

\(ab+\frac{1}{ab}\ge2\sqrt{ab.\frac{1}{ab}}=2\sqrt{1}=2\)

\(\Rightarrow ab+\frac{1}{ab}\ge2\)

\(0< a;b< 1\) thì không tìm được GTNN

\(gt\Rightarrow\left(2z-1\right)^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\) \(=6+3\left(\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right).\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(\Leftrightarrow\left(2z-1\right)^3=6+3.\sqrt[3]{9-8}\left(2z-1\right)=6z+3\)

\(\Leftrightarrow\left(2z-1\right)^3-6z-3=0\Leftrightarrow8z^3-12z^2-4=0\)

=> Đa thức bậc 3 có 1 nghiệm là z là:\(8x^3-12x^2-4=0\Leftrightarrow2x^3-3x^2-1=0\)

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

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

Để A=-1 thì \(-\sqrt{x-2}=-1\Leftrightarrow\sqrt{x-2}=1\)

\(\Leftrightarrow x-2=1\Rightarrow x=3\)

\(\sqrt{10\left(x-3\right)}=\sqrt{26}\)

\(\Rightarrow10\left(x-3\right)=26\)

\(\Rightarrow x-3=2.6\)

\(\Rightarrow x=3+2,6=5,6\)

\(\sqrt{3x^2}=x+2\Rightarrow3x^2=x^2+4x+4\)

\(\Rightarrow3x^2-x^2-4x-4=0\)

\(\Rightarrow2x^2-4x-4=0\)

\(\Rightarrow x^2-2x-2=0\)

\(a=1;b=-2;c=-2;b'=-1\)

\(\Delta'=b'^2-ac=\left(-1\right)^2-1.\left(-2\right)=3>0\)

Phương trình có 2 nghiệp phân biệt 

\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-\left(-1\right)+\sqrt{3}}{1}=1+\sqrt{3}\)

\(x_2=\frac{-b-\sqrt{\Delta'}}{a}=\frac{-\left(-1\right)-\sqrt{3}}{1}=1-\sqrt{3}\)

\(\sqrt{x^2+6x+9}=3x-6\)

\(x^2+6x+9=9x^2-36x+36\)

\(9x^2-x^2-36x-6x+36-9=0\)

\(8x^2-42x+27=0\)

\(a=8;b=-42;c=27;b'=-21\)

\(\Delta'=b'^2-ac=\left(-21\right)^2-8.27=225>0\)

Phương trình có 2 nghiệp phân biệt 

\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-\left(-21\right)+\sqrt{225}}{8}=\frac{21+15}{8}=\frac{36}{8}=\frac{9}{2}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-\left(-21\right)-\sqrt{225}}{8}=\frac{21-15}{8}=\frac{6}{8}=\frac{3}{4}\)

Ta có :

\(\(a^2+b^2+c^2=3\ge\frac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)\)

+) \(\(\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}=\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2b^2c^2}+\frac{4c^4}{2c^3+2c^2a^2}\)\)

\(\(\ge\frac{4\left(a^2+b^2+c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2b^2c^2+2c^2a^2}\)\)

\(\(\ge\frac{4.3^2}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2b^2c^2+2c^2a^2}\)\)

\(\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=\frac{36}{9+3}=3\ge a+b+c\left(dpcm\right)\)\)

_Minh ngụy_

Dễ thấy 

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

\(\Rightarrow a+b+c\le3\)

Do đó : 

\(\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}=\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2b^2c^2}+\frac{4c^4}{2c^3+2c^2a^2}\)

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

\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2b^2c^2+2c^2a^2}\)

\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\left(dpcm\right)\)

Lời giải :

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

Đặt \(\hept{\begin{cases}\sqrt{x-2}=a\\\sqrt{x+1}=b\end{cases}}\)

Ta có : \(b^2-a^2=x+1-x+2=3\)

Mặt khác : \(a+b=3\)( theo giả thiết )

Ta có : \(b^2-a^2=a+b\)

\(\Leftrightarrow\left(b-a\right)\left(a+b\right)=a+b\)

\(\Leftrightarrow b-a=1\)

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

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

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

\(\Leftrightarrow2\sqrt{x-2}=2\)

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

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow x=3\)( thỏa )

Vậy....