Question

[tran xuân phương]

Giải PT : \(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge-3\)

Đặt \(\sqrt{\frac{x+3}{2}}=a+1\ge0\Rightarrow x+3=2a^2+4a+2\)

Ta được hệ: \(\left\{{}\begin{matrix}2x^2+4x-a=1\\2a^2+4a-x=1\end{matrix}\right.\)

Trừ vế cho vế:

\(2\left(x^2-a^2\right)+4\left(x-a\right)+\left(x-a\right)=0\)

\(\Leftrightarrow\left(x-a\right)\left(2x+2a+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=x\\2\left(a+1\right)=-2x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{\frac{x+3}{2}}=x+1\\2\sqrt{\frac{x+3}{2}}=-2x-3\end{matrix}\right.\) \(\Leftrightarrow...\)

Answer 2

[le anh nhat]

ta có \(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

⇔4x⁴+16x³+16x²=\(\frac{x+3}{2}\)

⇔x+3=8x⁴+32x³+32x²

⇔x+3-8x⁴-32x³-32x²=0

⇔10x-9x+3-8x⁴-12x³-20x³+4x²-30x²-6x²=0

⇔(-6x²-9x+3)+(-8x⁴-12x³+4x²)+(-20x³-30x²+10x)

⇔-3(2x²+3x-1)-4x²(2x²+3x-1)-10x(2x²+3x-1)

⇔-(2x²+3x-1)(4x²+10x+3)

⇔\(\left[{}\begin{matrix}2x^2+3x-1=0\\4x^2+10+3=0\end{matrix}\right.\)

1. 2x²+3x-1=0

⇔x²+\(\frac{3}{2}\)x-\(\frac{1}{2}\)=0

⇔(x+\(\frac{3}{4}\))²=\(\frac{17}{16}\)

⇔\(x=\left\{{}\begin{matrix}\frac{-3+\sqrt{17}}{4}\\\frac{-3-\sqrt{17}}{4}\end{matrix}\right.\)

2.tương tự

x= \(\left\{{}\begin{matrix}\frac{-5-\sqrt{13}}{4}\\\frac{-5+\sqrt{13}}{4}\end{matrix}\right.\)

thử lại nghiệm thì chỉ có \(\frac{-3+\sqrt{17}}{4}\) và\(\frac{-5-\sqrt{13}}{4}\)thỏa mãn

⇒x=\(\frac{-3+\sqrt{17}}{4}\) và x=\(\frac{-5-\sqrt{13}}{4}\)

hơi dài