\(\dfrac{x+3}{x-3}+\dfrac{x-2}{x}=2\left(x\ne0;x\ne3\right)\\ \Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x\left(x-3\right)}{x\left(x-3\right)}\\ \Leftrightarrow x\left(x+3\right)+\left(x-2\right)\left(x-3\right)=2x\left(x-3\right)\\ \Leftrightarrow x^2+3x+x^2-5x+6=2x^2-6x\\ \Leftrightarrow2x^2-2x+6-2x^2+6x=0\\ \Leftrightarrow4x+6=0\\ \Leftrightarrow4x=-6\\ \Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
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\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\left(x\ne0\right)\\ 0\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\\ \Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\\ \Leftrightarrow2x^2-12-2x^2-3x\\ \Leftrightarrow-3x-12=0\\ \Leftrightarrow-3x=12\\ \Leftrightarrow x=\dfrac{12}{-3}\\ \Leftrightarrow x=-4\left(tm\right)\)
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\(\dfrac{x}{2x-1}=\dfrac{x-2}{2x+5}\left(x\ne\dfrac{1}{2};x\ne-\dfrac{5}{2}\right)\\ \Leftrightarrow x\left(2x+5\right)=\left(2x-1\right)\left(x-2\right)\\ \Leftrightarrow2x^2+5x=2x^2-4x-x+2\\ \Leftrightarrow2x^2+5x-2x^2+5x-2=0\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\left(tm\right)\)
các bạn nếu rõ điều kiện xác định và mẫu thức chung giúp mình nha^^
