\(a,PT\Leftrightarrow\dfrac{3\left(x+1\right)}{6}-\dfrac{2x}{6}=\dfrac{3}{6}\)
\(\Leftrightarrow3x+3-2x=3\)
\(\Leftrightarrow x=0\)
Vậy: \(S=\left\{0\right\}\)
\(b,PT\Leftrightarrow\left(x+5\right)\left(-6-x\right)\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)
Vậy: \(S=\left\{-5;-6\right\}\)
\(c,\) ĐKXĐ: \(x\ne\pm1\)
\(PT\Leftrightarrow\left(x-2\right)\left(x-1\right)-x\left(x+1\right)=6-x\)
\(\Leftrightarrow-4x+2=6-x\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
Vậy: \(S=\left\{-\dfrac{4}{3}\right\}\)
a, \(\Rightarrow3x+3-2x=3\Leftrightarrow x=0\)
b, \(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow\left(x+5\right)\left(-x-6\right)=0\Leftrightarrow x=-5;x=-6\)
c, đk : x khác -1 ; 1
\(\Rightarrow\left(x-2\right)\left(x-1\right)-x\left(x+1\right)=6-x\)
\(\Leftrightarrow x^2-3x+2-x^2-x=6-x\Leftrightarrow-4x+2=6-x\Leftrightarrow-3x=4\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)
