ĐKXĐ: \(x\ge-2\)
\(2\left(x^2-x+6\right)=5\sqrt{x^3+8}\)
\(\Leftrightarrow2\left(x^2-x+6\right)=5\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{x^2-2x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}=2\sqrt{x^2-2x+4}\\2\sqrt{x+2}=\sqrt{x^2-2x+4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=4\left(x^2-2x+4\right)\\4\left(x+2\right)=x^2-2x+4\end{matrix}\right.\)
\(\Rightarrow...\)
ĐKXĐ: x≥−2x≥−2
2(x2−x+6)=5√x3+82(x2−x+6)=5x3+8
⇔2(x2−x+6)=5√(x+2)(x2−2x+4)⇔2(x2−x+6)=5(x+2)(x2−2x+4)
Đặt {√x+2=a≥0√x2−2x+4=b>0{x+2=a≥0x2−2x+4=b>0
⇒2(a2+b2)=5ab⇒2(a2+b2)=5ab
⇔2a2−5ab+2b2=0⇔2a2−5ab+2b2=0
⇔(a−2b)(2a−b)=0⇔(a−2b)(2a−b)=0
⇒[a=2b2a=b⇒[a=2b2a=b ⇒[√x+2=2√x2−2x+42√x+2=√x2−2x+4⇒[x+2=2x2−2x+42x+2=x2−2x+4
⇒[x+2=4(x2−2x+4)4(x+2)=x2−2x+4⇒[x+2=4(x2−2x+4)4(x+2)=x2−2x+4
⇒...
