Question

\(\sqrt{4x+1}-\sqrt{5-2x}+2x^2-5x=0\)

Answer 1

ĐK \(\begin{cases}4x+1\ge0\\5-2x\ge0\end{cases}\Leftrightarrow }\frac{-1}{4}\le x\le\frac{5}{2}}\)
pt\(\Leftrightarrow\)\(\sqrt{4x+1}-3-\sqrt{5-2x}+1+2x^2-5x+2=0\)
\(\Leftrightarrow\frac{4x-8}{\sqrt{4x+1}+3}+\frac{2x-4}{\sqrt{5-2x}+1}+\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{4}{\sqrt{4x+1}+3}+\frac{2}{\sqrt{5-2x}+1}+2x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\\frac{4}{\sqrt{4x+1}+3}+\frac{2}{\sqrt{5-2x}+1}+2x-1=0\left(Vn_o\right)\end{array}\right.\)
KL: x=2