Question

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Answer 1

\(a,3\left(x-1\right)-4=2\left(x+1\right)-7\\ \Leftrightarrow3x-3-4=2x+2-7\\ \Leftrightarrow3x-7=2x-5\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ b,\left(x+1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{3}{x^2-9}\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{\left(x^2+6x+9\right)-\left(x^2-6x+9\right)-3}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow x^2+6x+9-x^2+6x-9-3=0\\ \Leftrightarrow12x-3=0\\ \Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

d, \(d,\dfrac{x-3}{4}-\dfrac{2x+3}{3}=\dfrac{5x-1}{3}+\dfrac{2x+9}{12}\\ \Leftrightarrow d,\dfrac{3\left(x-3\right)}{12}-\dfrac{4\left(2x+3\right)}{12}=\dfrac{4\left(5x-1\right)}{12}+\dfrac{2x+9}{12}\\ \Leftrightarrow3x-9-8x-12=20x-4+2x+9\\ \Leftrightarrow-5x-21=22x+5\\ \Leftrightarrow27x+26=0\\ \Leftrightarrow x=-\dfrac{26}{27}\left(tm\right)\)