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a, \(\dfrac{2x-3}{-5}>\dfrac{x-2}{-3}\)
<=> \(\dfrac{2x-3}{-5}.-15< \dfrac{x-2}{-3}.-15\)
<=> 3(2x - 3) < 5(x - 2)
<=> 6x - 9 < 5x - 10
<=> x < -1 S = {x|x<-1}
b, \(\dfrac{x-2}{6}-\dfrac{x-1}{3}\le\dfrac{x}{2}\)
<=> \(\dfrac{x-2}{6}-\dfrac{2x-2}{6}\le\dfrac{3x}{6}\)
<=> x - 2 - 2x + 2 \(\le\) 3x
<=> -x\(\le\) 3x
<=> 2x \(\le\) 0
<=> x \(\le\) 0 S = {x|x\(\le\)0}
c,\(2+\dfrac{3\left(x+1\right)}{3}< 3-\dfrac{x-1}{4}\)
<=> 2 + x + 1 < 3 - \(\dfrac{x-1}{4}\)
<=> 12 + x < 12 - x + 1
<=> 2x < 1
<=> x < \(\dfrac{1}{2}\) S = {x|x<\(\dfrac{1}{2}\)}
d,\(5+\dfrac{x+4}{5}< x-\dfrac{x-2}{2}+\dfrac{x+3}{3}\)
<=> \(\dfrac{150}{30}+\dfrac{6x+24}{30}< \dfrac{30x}{30}-\dfrac{15x-30}{30}+\dfrac{10x+30}{30}\)
<=> 150 + 6x + 24 < 30x - 15x + 30 + 10x + 30
<=> 114 < 19x
<=> x > 6 S = {x|x>6}
a) \(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
b) \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\\ \Leftrightarrow-2\left(2x-5\right)=0\\ \Leftrightarrow2x-5=0\\ \Leftrightarrow x=\dfrac{5}{2}\)
c) \(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)
\(\Leftrightarrow x^3+64-x^3+25x=264\)
\(\Leftrightarrow25x=200\)
hay x=8
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