\(a,5-\left(2-x\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-2+x=12-8x\)
\(\Leftrightarrow3+x=12-8x\)
\(\Leftrightarrow9x=9\)
\(\Leftrightarrow x=1\)
\(b,\left(x-4\right)\left(2x-3\right)=x^2-16\)
\(\Leftrightarrow\left(x-4\right)\left(2x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-3-x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
\(c,\dfrac{2}{x^2-6x+8}=\dfrac{3x-1}{x-2}-\dfrac{3x}{x-4}\left(dkxd:x\ne2;x\ne4\right)\)
\(\Leftrightarrow\dfrac{2-\left(3x-1\right)\left(x-4\right)+3x\left(x-2\right)}{x^2-6x+8}=0\)
\(\Leftrightarrow2-3x^2+12x+x-4+3x^2-6x=0\)
\(\Leftrightarrow7x=2\)
\(\Leftrightarrow x=\dfrac{2}{7}\left(tmdk\right)\)
Vậy ...
a) \(5-\left(2-x\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-2+x=12-8x\)
\(\Leftrightarrow3+x=12-8x\)
\(\Leftrightarrow9x=9\)
\(\Leftrightarrow x=\dfrac{9}{9}=1\)
b) \(\left(x-4\right)\left(2x-3\right)=x^2-16\)
\(\Leftrightarrow\left(x-4\right)\left(2x-3\right)=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(2x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
c) \(\dfrac{2}{x^2-6x+8}=\dfrac{3x-1}{x-2}-\dfrac{3x}{x-4}\)
\(\Leftrightarrow\dfrac{2}{x^2-6x+8}=\dfrac{\left(3x-1\right)\left(x-4\right)-3x\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Leftrightarrow\dfrac{2}{x^2-6x+8}=\dfrac{\left(3x-1\right)\left(x-4\right)-3x\left(x-2\right)}{x^2-4x-2x+8}\)
\(\Leftrightarrow2=3x^2-12x-x+4-3x^2+6x\)
\(\Leftrightarrow2=-7x+4\)
\(\Leftrightarrow7x=2\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
