Question

Giải phương trình:

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

Answer 1

ĐKXĐ:...

\(\Leftrightarrow4x^2-5x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x^2-3x+x^2-2x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)^2=-2\sqrt{x-1}\)

Đặt \(\sqrt{x-1}=t\left(t\ge0\right)\Rightarrow t^2=x-1\Leftrightarrow x=t^2+1\)

\(\Leftrightarrow3\left(t^2+1\right)t^2+t^4=-2t\)

\(\Leftrightarrow3t^4+3t^2+t^4+2t=0\)

\(\Leftrightarrow4t^4+3t^2+2t=0\)

\(\Leftrightarrow t\left(4t^3+3t+2\right)=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow t=0\Leftrightarrow x=1\left(tm\right)\)

Vậy x=1