Question

Giải phương trình \(3x^2+2x=2\sqrt{x^2+x}+1-x\)

Answer 1

ĐK: \(x\ge0\) hoặc \(x\le-1\)

PT ⇔ \(3x^2+3x-2\sqrt{x^2+x}-1=0\)

⇔ \(3\left(x^2+x\right)-3\sqrt{x^2+x}+\sqrt{x^2+x}-1=0\)

⇔ \(\left(\sqrt{x^2+x}-1\right)\left(3\sqrt{x^2+x}+1\right)=0\)

Do \(3\sqrt{x^2+x}+1>0\) nên \(x^2+x=1\)

⇔ \(x=\frac{\pm\sqrt{5}-1}{2}\) (t/m)

Vậy ...

Answer 2

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

Điều kiện: \(0\le x\le-1\)

\(\begin{array}{l} \Leftrightarrow - 2\sqrt {{x^2} + x} = 1 - x - 3{x^2} - 2x\\ \Leftrightarrow - 2\sqrt {{x^2} + x} = 1 - 3x - 3{x^2}\\ \Leftrightarrow 3{x^2} + 3x - 2\sqrt {{x^2} + x} - 1 = 0\\ \Leftrightarrow 3\left( {{x^2} + x} \right) + \sqrt {{x^2} + x} - 1 - 3\sqrt {{x^2} + x} = 0\\ \Leftrightarrow \sqrt {{x^2} + x} \left( {3\sqrt {{x^2} + x} + 1} \right) - \left( {3\sqrt {{x^2} + x} + 1} \right) = 0\\ \Leftrightarrow \left( {3\sqrt {{x^2} + x + 1} } \right)\left( {\sqrt {{x^2} + x }- 1 } \right) = 0 \end{array}\)

Ta thấy: \(3\sqrt{x^2+x+1}>0\Leftrightarrow x^2+x=1\)

\(\sqrt{x^2+x}-1=0\\ \Leftrightarrow x^2+x-1=0\\ \Leftrightarrow x=\frac{-1\pm\sqrt{5}}{2}\)