a/ ĐK: \(3x^2-10x+6\ge0\)
Nhận thấy \(x=0\) không phải nghiệm
\(\Leftrightarrow2\left(x^2+4\right)=\left(3x^2-10x+6\right)^2\)
\(\Leftrightarrow2\left(x^2+\frac{4}{x^2}\right)=\left(3x-10+\frac{6}{x}\right)^2=\left(3\left(x+\frac{2}{x}\right)-10\right)^2\)
Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)
\(\Leftrightarrow2\left(a^2-4\right)=\left(3a-10\right)^2\)
\(\Leftrightarrow7a^2-60a+108=0\Rightarrow\left[{}\begin{matrix}a=6\\a=\frac{18}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=6\\x+\frac{2}{x}=\frac{18}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+2=0\\7x^2-18x+14=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3+\sqrt{7}\\x=3-\sqrt{7}\end{matrix}\right.\)
b/ \(x\ge-\frac{1}{4}\)
Đặt \(\sqrt{x+\frac{1}{4}}=a\ge0\Rightarrow x=a^2-\frac{1}{4}\)
\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{a^2-\frac{1}{4}+\frac{1}{2}+a}=2\)
\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{\frac{1}{4}\left(4a^2+4a+1\right)}=2\)
\(\Leftrightarrow a^2-\frac{1}{4}+\frac{1}{2}\left(2a+1\right)=2\)
\(\Leftrightarrow4a^2+4a-7=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{-1+2\sqrt{2}}{2}\\a=\frac{-1-2\sqrt{2}}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+\frac{1}{4}}=\frac{-1+2\sqrt{2}}{2}\Rightarrow x=2-\sqrt{2}\)
d/ ĐKXĐ: \(\left|x\right|>3\)
\(x+\frac{3x}{\sqrt{x^2-9}}=6\sqrt{2}\)
Đặt \(\sqrt{x^2-9}=a\Rightarrow9=x^2-a^2\)
\(\Leftrightarrow x+\frac{3x}{a}=6\sqrt{2}\)
\(\Leftrightarrow6\sqrt{2}a-ax-3x=0\)
\(\Leftrightarrow6\sqrt{2}a-ax-3x+9\sqrt{2}-9\sqrt{2}=0\)
\(\Leftrightarrow6\sqrt{2}a-ax-3x+9\sqrt{2}-\sqrt{2}\left(x^2-a^2\right)=0\)
\(\Leftrightarrow\sqrt{2}a^2+\left(6\sqrt{2}-x\right)a-\sqrt{2}x^2-3x+9\sqrt{2}=0\)
\(\Delta=\left(6\sqrt{2}-x\right)^2-4\sqrt{2}\left(-\sqrt{2}x^2-3x+9\sqrt{2}\right)=9x^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{x-6\sqrt{2}+3x}{2\sqrt{2}}=\sqrt{2}x-3\\a=\frac{x-6\sqrt{2}-3x}{2\sqrt{2}}=\frac{-x-3\sqrt{2}}{\sqrt{2}}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-9}=\sqrt{2}x-3\\\sqrt{2x^2-18}=-x-3\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=\left(\sqrt{2}x-3\right)^2\left(x\ge\frac{3\sqrt{2}}{2}\right)\\2x^2-18=\left(-x-3\sqrt{2}\right)^2\left(x\le-3\sqrt{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6\sqrt{2}x+18=0\\x^2-6\sqrt{2}x-36=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\sqrt{2}\\x=3\sqrt{6}+3\sqrt{2}>-3\sqrt{2}\left(l\right)\\x=-3\sqrt{6}+3\sqrt{2}>-3\sqrt{2}\left(l\right)\end{matrix}\right.\)
Câu d còn 1 cách giải khác có vẻ nhẹ nhàng hơn 1 chút:
Nếu \(x< 0\Rightarrow VT=x\left(1+\frac{3}{\sqrt{x^2-9}}\right)< 0< VP\Rightarrow\) vô nghiệm
\(\Rightarrow x\ge0\Rightarrow VT\ge0\), bình phương 2 vế:
\(x^2+\frac{9x^2}{x^2-9}+2\frac{3x^2}{\sqrt{x^2-9}}=72\)
\(\Leftrightarrow\frac{x^4}{x^2-9}+6\frac{x^2}{\sqrt{x^2-9}}-72=0\)
Đặt \(\frac{x^2}{\sqrt{x^2-9}}=a\ge0\) ta được:
\(a^2+6a-72=0\Rightarrow\left[{}\begin{matrix}a=6\\a=-12< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\frac{x^2}{\sqrt{x^2-9}}=6\Leftrightarrow x^4=36\left(x^2-9\right)\)
\(\Leftrightarrow x^4-36x^2+324=0\Rightarrow x^2=18\Rightarrow x=3\sqrt{2}\)
