Question

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

Answer 1

ĐK:....

Đặt \(\sqrt{\frac{3x-1}{x}}=a\ge0\Leftrightarrow\frac{1}{a^2}=\frac{x}{3x-1}\)

\(pt\Leftrightarrow2a=\frac{1}{a^2}+1\)

\(\Leftrightarrow2a=\frac{a^2+1}{a^2}\)

\(\Leftrightarrow2a^3-a^2-1=0\)

\(\Leftrightarrow2a^3-2a^2+a^2-1=0\)

\(\Leftrightarrow2a^2\left(a-1\right)+\left(a-1\right)\left(a+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Leftrightarrow a=1\) ( vì \(2a^2+a+1>0\forall a\) )

Hay \(\sqrt{\frac{3x-1}{x}}=1\)

\(\Leftrightarrow\frac{3x-1}{x}=1\)

\(\Leftrightarrow3x-1=x\)

\(\Leftrightarrow x=\frac{1}{2}\) ( thỏa )

Vậy...