1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(ĐK:x\ne3;-2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)
\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)
\(\Leftrightarrow\) \(\left(x+1\right)^2+x-1=x^2+4\) hoặc \(\left(x+1\right)^2+1-x=x^2+4\)
Xét \(\left(x+1\right)^2+x-1=x^2+4\)
\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)
\(\Leftrightarrow3x-4=0\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
Xét \(\left(x+1\right)^2+1-x=x^2+4\)
\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)
2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)
\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)
\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)
\(\Leftrightarrow-x< 4\)
\(\Leftrightarrow x>4\)
Vậy \(S=\left\{x|x>4\right\}\)
