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Đặt \(x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, \Rightarrow x+y+z=2\)

Suy ra    \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)

Ta có \(\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{\left(2-x\right)^2} .\frac{2-x}{8}.\frac{2-x}{8}}=\frac{3x}{4}.\)

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\)\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge x+y+z-\frac{3}{2}=2-\frac{3}{2}=\frac{1}{2}\)

dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)hay \(a=b=c=\frac{3}{2}\)

  Đáp án

   Ta có: 

\(PT\Leftrightarrow\left|2x\right|+5=\sqrt{\left(2x-5\right)^2}=\left|2x-5\right|.\)

nếu x<0 thì PT \(\Leftrightarrow-2x+5=-2x+5.\)đúng với mọi x<0

nếu \(0\le x< \frac{5}{2}\)thì \(PT\Leftrightarrow2x+5=-2x+5x\Leftrightarrow x=0.\)

nếu \(x>\frac{5}{2}\Rightarrow PT\Leftrightarrow2x+5=2x-5\)vô nghiệm

Vậy nghiệm của phương trình là \(x\le0\)