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Đặt \(\sqrt{x+8}=y-3\Rightarrow x+8=y^2-6y+9\)
\(\Rightarrow x=y^2-6y+1\)
Ta được hệ:
\(\left\{{}\begin{matrix}x^2-6x-2=y-3\\x=y^2-6y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-6x+1=y\\y^2-6x+1=x\end{matrix}\right.\)
Trừ vế:
\(\Rightarrow x^2-y^2-5x+5y=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-5\right)=0\Rightarrow\left[{}\begin{matrix}x=y\\y=5-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3+\sqrt{x+8}\\3+\sqrt{x+8}=5-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+8}=x-3\left(x\ge3\right)\\\sqrt{x+8}=2-x\left(x\le2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=x^2-6x+9\left(x\ge3\right)\\x+8=x^2-4x+4\left(x\le2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-7x+1=0\left(x\ge3\right)\\x^2-5x-4=0\left(x\le2\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{5-\sqrt{41}}{2}\end{matrix}\right.\)
\(x^2-6x-2=\sqrt{x+8}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-6x-2\ge0\\\left(x^2-6x-2\right)^2=\left(\sqrt{x+8}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3-\sqrt{5}\cup x\ge3+\sqrt{5}\left(1\right)\\x^4-12x^3+32x^2+24x+4=x+8\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^4-12x^3+32x^2+23x-4=0\)
\(\Leftrightarrow\left(x^4-7x^3+x^2\right)+\left(-5x^3+35x^2-5x\right)+\left(-4x^2+28x-4\right)=0\)
\(\Leftrightarrow x^2\left(x^2-7x+1\right)-5x\left(x^2-7x+1\right)-4\left(x^2-7x+1\right)=0\)
\(\Leftrightarrow\left(x^2-7x+1\right)\left(x^2-5x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-7x+1=0\\x^2-5x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7\pm3\sqrt{5}}{2}\\x=\dfrac{5\pm\sqrt{41}}{2}\end{matrix}\right.\) so với điều kiện \(\left(1\right)\) thỏa
Vậy phương trình cho có \(4\) nghiệm \(x\in\left(\dfrac{7\pm3\sqrt{5}}{2};\dfrac{5\pm\sqrt{41}}{2}\right)\)
Kiểm tra lại bằng đồ thị :
\(y=x^4-12x^3+32x^2+23x-4\)
Ta thấy đồ thị hàm số cắt trục hoành tại 4 điểm phân biệt
