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ĐKXĐ: ...
\(\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)
\(\Leftrightarrow\frac{2}{x+\frac{12}{x}-3}+\frac{6}{x+\frac{12}{x}+2}=1\)
Đặt \(x+\frac{12}{x}-3=t\)
\(\Rightarrow\frac{2}{t}+\frac{6}{t+5}=1\Leftrightarrow2\left(t+5\right)+6t=t\left(t+5\right)\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{12}{x}-3=-2\\x+\frac{12}{x}-3=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+12=0\\x^2-8x+12=0\end{matrix}\right.\) (casio)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.ĐK: 2x2+1\(\ne0\) \(\forall x\)
Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x2+1 >0 với mọi x)
\(\Leftrightarrow x=2\) (TM)
Vậy ...
b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)
Để phương trình bằng 0 thì x2-x-6=0 (Vì x-3\(\ne0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)
Vậy ...
c. ĐK: x\(\ne\)2
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)
\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)
\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)
Vậy ...
d. ĐK: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)
\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))
\(\Leftrightarrow x=-2\:\left(TM\right)\)
Vậy...
\(pt\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)
\(\Leftrightarrow\frac{2}{x-3+\frac{12}{x}}+\frac{6}{x+2+\frac{12}{x}}=1\)
Đặt \(x+\frac{12}{x}=t\)
Khi đó:
\(pt\Leftrightarrow\frac{2}{t-3}+\frac{6}{t+2}=1\Leftrightarrow2t+4+6t-18=t^2-t-6\)
\(\Leftrightarrow t^2-t-6=8t-14\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left(t-8\right)\left(t-1\right)=0\)
\(\Leftrightarrow x+\frac{12}{x}=8;x+\frac{12}{x}=1\)
Thôi,bí rồi