=2023

2023 , kb nha 

\(ĐK:4x-1\ge0\Leftrightarrow x\ge\frac{1}{4}\)

\(pt\Leftrightarrow\frac{x}{\sqrt{4x-1}}-2+\frac{\sqrt{4x-1}}{x}=0\)

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

\(\Rightarrow x=\sqrt{4x-1}\Rightarrow x^2=4x-1\Leftrightarrow x^2-4x+1=0\)

\(\Leftrightarrow\left(x-2\right)^2=3\Rightarrow\orbr{\begin{cases}x-2=\sqrt{3}\\x-2=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)

Nguyễn Hưng Phát ĐKXĐ : \(x>\frac{1}{4}\) mới đúng nha nhok :v 

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)

Theo đề bài \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=0\)

\(\Rightarrow\frac{c+a+b}{abc}=0\) mà \(a;b;c\ne0\Rightarrow abc\ne0\Rightarrow a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=-\left(a^3+b^3+c^3\right)\)

Mà \(3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) chia hết cho 3 nên \(-\left(a^3+b^3+c^3\right)\) chia hết cho 3

Nên \(a^3+b^3+c^3\) chia hết cho 3